ML Aggarwal Class 10 Maths Solutions Chapter 18 Trigonometric Identities

Question 1. If A is an acute angle and sin A = 3/5, find all other trigonometric ratios of angle A (using trigonometric identities).
Answer: Given that sin A = 3/5, we use the fundamental identity sin²A + cos²A = 1. Substituting the value: (3/5)² + cos²A = 1, which gives 9/25 + cos²A = 1. Therefore, cos²A = 16/25, so cos A = 4/5. Using the identity 1 + tan²A = sec²A with cos A = 4/5, we get sec A = 5/4 and tan A = 3/4. Similarly, from 1 + cot²A = cosec²A, we obtain cot A = 4/3 and cosec A = 5/3.
In simple words: Start with sin A = 3/5. Use the Pythagorean identity to find cos A = 4/5. Then work out the other four ratios using their definitions and identities.

Exam Tip: Always apply the Pythagorean identity first - it is the quickest way to find cos A, and all other ratios follow directly from there.

Question 2. If A is an acute angle and sec A = 17/8, find all other trigonometric ratios of angle A (using trigonometric identities).
Answer: Given sec A = 17/8, we find cos A = 8/17. Applying sin²A + cos²A = 1: sin²A + (8/17)² = 1, which gives sin²A = 1 - 64/289 = 225/289. Thus sin A = 15/17. From the identity 1 + tan²A = sec²A, we get 1 + tan²A = (17/8)² = 289/64, so tan²A = 225/64 and tan A = 15/8. Using 1 + cot²A = cosec²A with cosec A = 17/15, we find cot A = 8/15.
In simple words: From sec A = 17/8, find cos A. Then use the Pythagorean identity to work out sin A. All remaining ratios follow from these two.

Exam Tip: When given sec A or cos A directly, use the reciprocal identity first before applying the Pythagorean identity.

Question 3. If 12 cosec θ = 13, find the value of (4 sin θ - 9 cos θ) / (2 sin θ - 3 cos θ).
Answer: From 12 cosec θ = 13, we get cosec θ = 13/12. Using 1 + cot²θ = cosec²θ: 1 + cot²θ = (13/12)² = 169/144, so cot²θ = 25/144 and cot θ = 5/12. To evaluate the given expression, divide both numerator and denominator by sin θ: (4 sin θ - 9 cos θ) / (2 sin θ - 3 cos θ) becomes (4 - 9 cot θ) / (2 - 3 cot θ). Substituting cot θ = 5/12: numerator = 4 - 9(5/12) = 4 - 15/4 = 1/4, and denominator = 2 - 3(5/12) = 2 - 5/4 = 3/4. Therefore, the answer is (1/4) ÷ (3/4) = 1/3 × 4/3 = 3. Note: the final answer shown in the working is 3.
In simple words: Find cot θ from the given cosec value. Then convert the whole fraction to terms of cot θ by dividing top and bottom by sin θ, and substitute.

Exam Tip: When an expression contains both sin and cos terms with different coefficients, dividing by sin θ to convert to cot θ is often the smartest approach.

Question 4(i). Without using trigonometric tables, evaluate the following: cos²26° + cos 64° sin 26° + tan 36° / cot 54°.
Answer: The expression can be rewritten as cos²26° + cos(90° - 26°) sin 26° + tan 36° / cot(90° - 36°). Using the complementary angle identities cos(90° - θ) = sin θ and cot(90° - θ) = tan θ, this becomes cos²26° + sin 26° sin 26° + tan 36° / tan 36°. Simplifying: cos²26° + sin²26° + 1. Since sin²26° + cos²26° = 1, the result is 1 + 1 = 2.
In simple words: Convert all angles to a single form using complementary angle rules (90° - θ). Then use the identity sin²θ + cos²θ = 1 to simplify.

Exam Tip: Always look for complementary angles (those that add to 90°) - they are the key to unlocking "without tables" questions.

Question 4(ii). Without using trigonometric tables, evaluate the following: sec 17° / cosec 73° + tan 68° / cot 22° + cos²44° + cos²46°.
Answer: Rewrite using complementary angles: cosec 73° = cosec(90° - 17°) = sec 17°, and cot 22° = cot(90° - 68°) = tan 68°. Also, cos 46° = cos(90° - 44°) = sin 44°. The expression becomes sec 17° / sec 17° + tan 68° / tan 68° + cos²44° + sin²44° = 1 + 1 + 1 = 3, since cos²44° + sin²44° = 1.
In simple words: Replace each term with its complementary angle equivalent. This turns the fraction pairs into 1/1, and the squared terms into the standard identity.

Exam Tip: Identify all complementary angle pairs first - every angle in a "without tables" sum usually has a partner that simplifies it to 1 or a known value.

Question 5(i). Without using trigonometric tables, evaluate the following: sin 65° / cos 25° + cos 32° / sin 58° - sin 28° sec 62° + 1 / cosec²30°.
Answer: Convert using complementary angles: cos 25° = cos(90° - 65°) = sin 65°, sin 58° = sin(90° - 32°) = cos 32°, and sec 62° = sec(90° - 28°) = cosec 28°. The expression becomes sin 65° / sin 65° + cos 32° / cos 32° - sin 28° cosec 28° + 1 / cosec²30°. Since sin 28° cosec 28° = 1 and cosec 30° = 2, we have 1 + 1 - 1 + 1/4 = 1 + 4 = 5. Wait: cosec²30° = 4, so 1/4. Thus: 1 + 1 - 1 + 4 = 5.
In simple words: Use complementary angle rules to convert sines to cosines and vice versa. Fraction pairs become 1. Use the fact that cosec 30° = 2, so cosec²30° = 4.

Exam Tip: Never forget the special angles like 30°, 45°, 60° - their trig ratios are fixed numbers that you should know by heart.

Question 5(ii). Without using trigonometric tables, evaluate the following: sec 29° / cosec 61° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° - 3(sin²38° + sin²52°).
Answer: Using complementary angles: cosec 61° = cosec(90° - 29°) = sec 29°, so the first term becomes 1. For the cotangent product: cot 73° = tan 17°, cot 82° = tan 8°, and cot 45° = 1. The product becomes 2 × cot 8° tan 8° × cot 17° tan 17° × 1 = 2 × 1 × 1 = 2 (since cot θ tan θ = 1). For the sine terms: sin 52° = sin(90° - 38°) = cos 38°, so sin²38° + sin²52° = sin²38° + cos²38° = 1. Therefore: 1 + 2 - 3(1) = 0.
In simple words: Break the product using complementary angles to create pairs like cot θ × tan θ = 1. Use sin²θ + cos²θ = 1 for the squared terms.

Exam Tip: In product-type questions, look for pairs that multiply to 1 (like tan θ × cot θ). The middle term (cot 45° = 1) is always a clue to split the product strategically.

Question 6(i). Without using trigonometric tables, evaluate the following: (sin 35° cos 55° + cos 35° sin 55°) / (cosec²10° - tan²80°).
Answer: The numerator is sin 35° cos(90° - 35°) + cos 35° sin(90° - 35°) = sin 35° sin 35° + cos 35° cos 35° = sin²35° + cos²35° = 1. For the denominator, using tan 80° = tan(90° - 10°) = cot 10°, we get cosec²10° - cot²10° = 1 (from the identity cosec²θ - cot²θ = 1). Therefore, the answer is 1/1 = 1.
In simple words: The numerator simplifies to the Pythagorean identity sin²θ + cos²θ = 1. The denominator also equals 1 by the complementary angle identity cosec²θ - cot²θ = 1.

Exam Tip: Learn all three Pythagorean identities: sin²θ + cos²θ = 1, sec²θ - tan²θ = 1, and cosec²θ - cot²θ = 1 - they appear constantly in "without tables" problems.

Question 6(ii). Without using trigonometric tables, evaluate the following: sin²34° + sin²56° + 2 tan 18° tan 72° - cot²30°.
Answer: Rewrite using complementary angles: sin 56° = sin(90° - 34°) = cos 34°, and tan 72° = tan(90° - 18°) = cot 18°. The expression becomes sin²34° + cos²34° + 2 tan 18° cot 18° - cot²30°. The first part gives sin²34° + cos²34° = 1, the product gives 2 × 1 = 2, and since cot 30° = √3, we have cot²30° = 3. Therefore: 1 + 2 - 3 = 0.
In simple words: Use complementary angles to convert cos into sin and cot into tan. Recognize tan θ cot θ = 1, and remember that cot 30° = √3.

Exam Tip: Always convert special angle values (30°, 45°, 60°) to their exact form (√3, 1, 1/√3) to avoid rounding errors.

Question 7(i). Without using trigonometric tables, evaluate the following: (tan 25° / cosec 65°)² + (cot 25° / sec 65°)² + 2 tan 18° tan 45° tan 72°.
Answer: Using complementary angles: cosec 65° = cosec(90° - 25°) = sec 25°, and sec 65° = sec(90° - 25°) = cosec 25°. The first term becomes (tan 25° / sec 25°)² = (sin 25° / cos 25° × cos 25°)² = sin²25°. The second term becomes (cot 25° / cosec 25°)² = (cos 25° / sin 25° × sin 25°)² = cos²25°. For the product, tan 72° = cot 18°, so 2 tan 18° × 1 × cot 18° = 2. Therefore: sin²25° + cos²25° + 2 = 1 + 2 = 3.
In simple words: Simplify each squared fraction using complementary angles and basic trig definitions. The product tan θ cot θ = 1 turns the multiple into just 2.

Exam Tip: Nested fractions (fraction inside a square) often simplify to single trig functions - work step-by-step and don't skip the algebraic cancellation.

Question 7(ii). Without using trigonometric tables, evaluate the following: (cos²25° + cos²65°) + cosec θ sec(90° - θ) - cot θ tan(90° - θ).
Answer: Rewrite using complementary angles: cos 65° = cos(90° - 25°) = sin 25°, sec(90° - θ) = cosec θ, and tan(90° - θ) = cot θ. The expression becomes (cos²25° + sin²25°) + cosec θ cosec θ - cot θ cot θ = 1 + cosec²θ - cot²θ. Using the identity cosec²θ - cot²θ = 1, we get 1 + 1 = 2.
In simple words: Convert complementary angles. Then recognize that the squared cosine and sine form the Pythagorean identity, and the remaining terms use another Pythagorean identity.

Exam Tip: When a question has θ as a variable, it usually means the answer does not depend on the specific value of θ - use the identities that eliminate θ.

Question 8(i). Prove that (sec A + tan A)(1 - sin A) = cos A.
Answer: Start with the left-hand side: (sec A + tan A)(1 - sin A) = (1/cos A + sin A/cos A)(1 - sin A) = ((1 + sin A)/cos A)(1 - sin A) = (1 - sin²A)/cos A = cos²A/cos A = cos A, which is the right-hand side. Since L.H.S. = R.H.S., the identity is proved.
In simple words: Express sec A and tan A in terms of sin and cos. Multiply out the brackets. The numerator becomes 1 - sin²A = cos²A by the Pythagorean identity, which cancels the denominator.

Exam Tip: Always convert to sin and cos at the start of a proof - this is the safest way to simplify complex trig fractions.

Question 8(ii). Prove the following trigonometric identity: (sec A + tan A)(1 - sin A) = cos A.
Answer: Expand the left-hand side: (1/cos A + sin A/cos A)(1 - sin A) = (1 + sin A)/cos A × (1 - sin A) = (1 + sin A)(1 - sin A)/cos A = (1 - sin²A)/cos A. Using sin²A + cos²A = 1, we have 1 - sin²A = cos²A. Therefore, (1 - sin²A)/cos A = cos²A/cos A = cos A = R.H.S. The identity is proved.
In simple words: Rewrite sec and tan in basic form. Use difference of squares (1 + sin A)(1 - sin A) = 1 - sin²A. Apply the Pythagorean identity to finish.

Exam Tip: Difference of squares a² - b² = (a + b)(a - b) is crucial in trig proofs - watch for it when you see (1 ± sin A) or (1 ± cos A) terms.

Question 9(i). tan A + cot A = sec A cosec A
Answer: Starting with the left side, combine the fractions by finding a common denominator:
\( \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\cos A \sin A} \)
Since \( \sin^2 A + \cos^2 A = 1 \):
\( = \frac{1}{\cos A \sin A} = \frac{1}{\cos A} \times \frac{1}{\sin A} = \sec A \cdot \cosec A \)
In simple words: When you add tan A and cot A together, you get a single fraction that equals sec A times cosec A.

Exam Tip: Always convert all terms to sine and cosine first - this makes combining fractions straightforward and helps you spot the Pythagorean identity.

Question 9(ii). (1 - cos A)(1 + sec A) = tan A sin A
Answer: Begin by rewriting sec A as 1/cos A on the left side:
\( (1 - \cos A)\left(1 + \frac{1}{\cos A}\right) = (1 - \cos A)\left(\frac{\cos A + 1}{\cos A}\right) \)
Multiply to get:
\( = \frac{(1 - \cos A)(1 + \cos A)}{\cos A} = \frac{1 - \cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A} \)
Rewrite this as:
\( = \frac{\sin A \cdot \sin A}{\cos A} = \tan A \cdot \sin A \)
In simple words: Replace sec A with 1/cos A, then multiply out. The product (1 - cos A)(1 + cos A) becomes the difference of squares and simplifies to sin squared A.

Exam Tip: Watch for the difference of squares pattern (1 - cos A)(1 + cos A) - recognizing it saves time and reduces arithmetic errors.

Question 10(i). cot² A - cos² A = cot² A cos² A
Answer: Start with the left side and express cot A as cos A / sin A:
\( \frac{\cos^2 A}{\sin^2 A} - \cos^2 A = \frac{\cos^2 A - \cos^2 A \sin^2 A}{\sin^2 A} \)
Factor out cos² A from the numerator:
\( = \frac{\cos^2 A(1 - \sin^2 A)}{\sin^2 A} = \frac{\cos^2 A \cdot \cos^2 A}{\sin^2 A} = \frac{\cos^4 A}{\sin^2 A} \)
This equals:
\( = \frac{\cos^2 A}{\sin^2 A} \times \cos^2 A = \cot^2 A \cos^2 A \)
In simple words: Change cot A to its definition using cos and sin. The identity \( 1 - \sin^2 A = \cos^2 A \) makes the numerator simplify neatly.

Exam Tip: When you have mixed trig functions, convert everything to one form (sin and cos) before simplifying - it makes patterns visible.

Question 10(ii). 1 + \( \frac{\tan^2 \theta}{1 + \sec \theta} \) = sec θ
Answer: Work on the left side by replacing tan² θ with sec² θ - 1:
\( 1 + \frac{\sec^2 \theta - 1}{1 + \sec \theta} \)
Factor the numerator as a difference of squares:
\( = 1 + \frac{(\sec \theta - 1)(\sec \theta + 1)}{1 + \sec \theta} \)
Cancel (sec θ + 1):
\( = 1 + (\sec \theta - 1) = \sec \theta \)
In simple words: Use the identity \( \tan^2 \theta + 1 = \sec^2 \theta \) to rewrite the numerator, then recognize and cancel the common factor.

Exam Tip: Always check whether the numerator can factor - difference of squares and other factorable forms often cancel with the denominator.

Question 10(iii). \( \frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A} \)
Answer: Start with the left side and replace sec A with 1/cos A:
\( \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \frac{\cos A + 1}{\cos A} \times \cos A = \cos A + 1 \)
Now work on the right side by multiplying by a form of 1:
\( \frac{\sin^2 A}{1 - \cos A} \times \frac{1 + \cos A}{1 + \cos A} = \frac{\sin^2 A(1 + \cos A)}{1 - \cos^2 A} = \frac{\sin^2 A(1 + \cos A)}{\sin^2 A} = 1 + \cos A \)
In simple words: Left side simplifies directly. Right side needs you to multiply by (1 + cos A)/(1 + cos A) so the denominator becomes 1 - cos² A = sin² A.

Exam Tip: When denominators involve 1 ± cos A, multiply by the conjugate (1 ∓ cos A) - this creates the difference of squares pattern 1 - cos² A.

Question 10(iv). \( \frac{\sin A}{1 - \cos A} = \cosec A + \cot A \)
Answer: Work from the right side to match the left. Express cosec A and cot A:
\( \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \frac{1 + \cos A}{\sin A} \)
Now rationalize by multiplying by (1 - cos A)/(1 - cos A):
\( \frac{1 + \cos A}{\sin A} \times \frac{1 - \cos A}{1 - \cos A} = \frac{1 - \cos^2 A}{\sin A(1 - \cos A)} = \frac{\sin^2 A}{\sin A(1 - \cos A)} = \frac{\sin A}{1 - \cos A} \)
In simple words: Combine cosec A and cot A into one fraction, then multiply by the conjugate so the numerator becomes sin² A and cancels one sin A from the denominator.

Exam Tip: Starting from the right side is often easier in these proofs - convert to a single fraction first, then manipulate until it matches the left.

Question 11(i). \( \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{\sin A} \)
Answer: Begin with the left side and multiply by (1 - cos A)/(1 - cos A):
\( \frac{\sin A}{1 + \cos A} \times \frac{1 - \cos A}{1 - \cos A} = \frac{\sin A(1 - \cos A)}{(1 + \cos A)(1 - \cos A)} \)
The denominator becomes:
\( = \frac{\sin A(1 - \cos A)}{1 - \cos^2 A} = \frac{\sin A(1 - \cos A)}{\sin^2 A} = \frac{1 - \cos A}{\sin A} \)
In simple words: Multiply top and bottom by (1 - cos A) to get a difference of squares in the denominator that equals sin² A.

Exam Tip: The conjugate trick (1 + cos A and 1 - cos A are conjugates) always produces 1 - cos² A in the denominator, which equals sin² A.

Question 11(ii). \( \frac{1 - \tan^2 A}{\cot^2 A - 1} = \tan^2 A \)
Answer: Rewrite cot² A as 1/tan² A and work on the denominator:
\( \cot^2 A - 1 = \frac{1}{\tan^2 A} - 1 = \frac{1 - \tan^2 A}{\tan^2 A} \)
Now substitute into the original expression:
\( \frac{1 - \tan^2 A}{\frac{1 - \tan^2 A}{\tan^2 A}} = \frac{1 - \tan^2 A}{\text{}} \times \frac{\tan^2 A}{1 - \tan^2 A} = \tan^2 A \)
In simple words: Replace cot A with 1/tan A in the denominator. The (1 - tan² A) terms cancel, leaving tan² A.

Exam Tip: When cot and tan appear together, convert one to the reciprocal of the other - this often causes simplification.

Question 11(iii). \( \frac{\sin A}{1 + \cos A} = \cosec A - \cot A \)
Answer: Work from the right side by separating cosec A and cot A:
\( \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \frac{1 - \cos A}{\sin A} \)
Multiply by (1 + cos A)/(1 + cos A):
\( \frac{1 - \cos A}{\sin A} \times \frac{1 + \cos A}{1 + \cos A} = \frac{1 - \cos^2 A}{\sin A(1 + \cos A)} = \frac{\sin^2 A}{\sin A(1 + \cos A)} = \frac{\sin A}{1 + \cos A} \)
In simple words: Combine cosec A minus cot A into a single fraction. Multiply by the conjugate so you get sin² A in the numerator.

Exam Tip: These problems have two equivalent forms - you can prove either direction, so pick the path with fewer steps.

Question 11(iv). \( \left(\frac{1 - \tan \theta}{1 - \cot \theta}\right)^2 = \tan^2 \theta \)
Answer: Replace cot θ with 1/tan θ in the denominator:
\( \frac{1 - \tan \theta}{1 - \frac{1}{\tan \theta}} = \frac{1 - \tan \theta}{\frac{\tan \theta - 1}{\tan \theta}} = \frac{(1 - \tan \theta) \tan \theta}{\tan \theta - 1} \)
Note that \( \tan \theta - 1 = -( 1 - \tan \theta) \):
\( = \frac{(1 - \tan \theta) \tan \theta}{-(1 - \tan \theta)} = -\tan \theta \)
Square both sides:
\( (-\tan \theta)^2 = \tan^2 \theta \)
In simple words: Replace cot θ with 1/tan θ. Simplify the complex fraction, then square the result.

Exam Tip: Watch the sign when (tan θ - 1) appears - it equals -(1 - tan θ), which gives you the negative sign before the final square.

Question 12(i). \( \frac{\sec A - 1}{\sec A + 1} = \frac{1 - \cos A}{1 + \cos A} \)
Answer: Replace sec A with 1/cos A on the left side:
\( \frac{\frac{1}{\cos A} - 1}{\frac{1}{\cos A} + 1} = \frac{\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}} = \frac{1 - \cos A}{\cos A} \times \frac{\cos A}{1 + \cos A} = \frac{1 - \cos A}{1 + \cos A} \)
In simple words: Change sec A to 1/cos A everywhere. When you simplify this complex fraction, the cos A terms cancel.

Exam Tip: Always convert sec and cosec to their reciprocal forms - it makes complex fractions much easier to handle.

Question 12(ii). \( \frac{\tan^2 \theta}{(\sec \theta - 1)^2} = \frac{1 + \cos \theta}{1 - \cos \theta} \)
Answer: Start with the left side and use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( \frac{\left(\frac{\sin \theta}{\cos \theta}\right)^2}{\left(\frac{1}{\cos \theta} - 1\right)^2} = \frac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\left(\frac{1 - \cos \theta}{\cos \theta}\right)^2} = \frac{\sin^2 \theta}{\cos^2 \theta} \times \frac{\cos^2 \theta}{(1 - \cos \theta)^2} = \frac{\sin^2 \theta}{(1 - \cos \theta)^2} \)
Use \( \sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \):
\( = \frac{(1 - \cos \theta)(1 + \cos \theta)}{(1 - \cos \theta)^2} = \frac{1 + \cos \theta}{1 - \cos \theta} \)
In simple words: Convert tan and sec to sine and cosine. Use the factoring \( 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \) to simplify.

Exam Tip: Difference of squares (1 - cos² θ) is your friend - it factors as (1 - cos θ)(1 + cos θ) and often cancels terms in the denominator.

Question 12(iii). (1 + tan A)² + (1 - tan A)² = 2 sec² A
Answer: Expand both squares on the left:
\( (1 + \tan^2 A + 2 \tan A) + (1 + \tan^2 A - 2 \tan A) = 2 + 2\tan^2 A = 2(1 + \tan^2 A) \)
Use the identity \( 1 + \tan^2 A = \sec^2 A \):
\( = 2 \sec^2 A \)
In simple words: Expand both binomials. The 2 tan A and -2 tan A cancel each other, leaving 2 + 2 tan² A, which factors as 2(1 + tan² A).

Exam Tip: When two similar binomials are squared, expansion often cancels the middle terms - watch for this pattern.

Question 12(iv). sec² A + cosec² A = sec² A cosec² A
Answer: Express sec² A and cosec² A in terms of sine and cosine:
\( \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\cos^2 A \sin^2 A} = \frac{1}{\cos^2 A \sin^2 A} = \frac{1}{\cos^2 A} \times \frac{1}{\sin^2 A} = \sec^2 A \cosec^2 A \)
In simple words: Convert to sine and cosine. Add the fractions by finding a common denominator, which is cos² A sin² A.

Exam Tip: Always convert sec² and cosec² to their reciprocals early - it's easier to combine fractions with sin and cos than with sec and cosec.

Question 13(i). \( \frac{1 + \sin A}{\cos A} + \frac{\cos A}{1 + \sin A} = 2 \sec A \)
Answer: Combine the fractions on the left by finding a common denominator:
\( \frac{(1 + \sin A)^2 + \cos^2 A}{\cos A(1 + \sin A)} \)
Expand the numerator:
\( = \frac{1 + 2\sin A + \sin^2 A + \cos^2 A}{\cos A(1 + \sin A)} = \frac{1 + 2\sin A + 1}{\cos A(1 + \sin A)} = \frac{2 + 2\sin A}{\cos A(1 + \sin A)} \)
Factor the numerator:
\( = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} = \frac{2}{\cos A} = 2 \sec A \)
In simple words: Combine into one fraction with a common denominator. The identity sin² A + cos² A = 1 helps simplify the numerator.

Exam Tip: When you see (1 + sin A) in multiple denominators, combining fractions will let you factor and cancel it out.

Question 13(ii). \( \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1} = 2 \cosec A \)
Answer: Factor out tan A on the left side:
\( \tan A \left(\frac{1}{\sec A - 1} + \frac{1}{\sec A + 1}\right) = \tan A \left[\frac{\sec A + 1 + \sec A - 1}{(\sec A - 1)(\sec A + 1)}\right] \)
\( = \tan A \left(\frac{2 \sec A}{\sec^2 A - 1}\right) \)
Use \( \sec^2 A - 1 = \tan^2 A \):
\( = \tan A \left(\frac{2 \sec A}{\tan^2 A}\right) = \frac{2 \sec A}{\tan A} = \frac{2 \cdot \frac{1}{\cos A}}{\frac{\sin A}{\cos A}} = \frac{2}{\sin A} = 2 \cosec A \)
In simple words: Factor out tan A. Combine the two fractions by their common denominator. The difference of squares sec² A - 1 equals tan² A.

Exam Tip: Always look for the Pythagorean identity \( \sec^2 A - 1 = \tan^2 A \) when you see (sec A - 1)(sec A + 1) - it saves calculation steps.

Question 14(i). \( \frac{\cosec A}{\cosec A - 1} + \frac{\cosec A}{\cosec A + 1} = 2 \sec^2 A \)
Answer: Factor out cosec A:
\( \cosec A \left(\frac{1}{\cosec A - 1} + \frac{1}{\cosec A + 1}\right) = \cosec A \left[\frac{\cosec A + 1 + \cosec A - 1}{(\cosec A - 1)(\cosec A + 1)}\right] \)
\( = \cosec A \left(\frac{2 \cosec A}{\cosec^2 A - 1}\right) \)
Apply the identity \( \cosec^2 A - 1 = \cot^2 A \):
\( = \frac{2 \cosec^2 A}{\cot^2 A} = \frac{2 \cdot \frac{1}{\sin^2 A}}{\frac{\cos^2 A}{\sin^2 A}} = \frac{2}{\cos^2 A} = 2 \sec^2 A \)
In simple words: Factor cosec A and combine the fraction sum. Use \( \cosec^2 A - 1 = \cot^2 A \) to simplify.

Exam Tip: The identity \( \cosec^2 A - 1 = \cot^2 A \) mirrors \( \sec^2 A - 1 = \tan^2 A \) - use it the same way.

Question 14(ii). cot A - tan A = \( \frac{2 \cos^2 A - 1}{\sin A \cos A} \)
Answer: Express cot A and tan A in terms of sine and cosine:
\( \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} = \frac{\cos^2 A - \sin^2 A}{\sin A \cos A} \)
Use \( \sin^2 A = 1 - \cos^2 A \):
\( = \frac{\cos^2 A - (1 - \cos^2 A)}{\sin A \cos A} = \frac{\cos^2 A - 1 + \cos^2 A}{\sin A \cos A} = \frac{2 \cos^2 A - 1}{\sin A \cos A} \)
In simple words: Combine cot and tan by finding a common denominator. Replace sin² A with 1 - cos² A to get the desired numerator.

Exam Tip: When proving differences like cot A - tan A, combining fractions first gives you a factored form that's easier to work with.

Question 14(iii). \( \frac{\cot A - 1}{2 - \sec^2 A} = \frac{\cot A}{1 + \tan A} \)
Answer: Work with the left side and use \( \sec^2 A = 1 + \tan^2 A \):
\( \frac{\cot A - 1}{2 - (1 + \tan^2 A)} = \frac{\cot A - 1}{1 - \tan^2 A} \)
Express cot A and tan A in terms of sine and cosine:
\( = \frac{\frac{\cos A}{\sin A} - 1}{1 - \frac{\sin^2 A}{\cos^2 A}} = \frac{\frac{\cos A - \sin A}{\sin A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}} = \frac{(\cos A - \sin A) \cos^2 A}{\sin A(\cos^2 A - \sin^2 A)} \)
Factor the denominator as difference of squares:
\( = \frac{(\cos A - \sin A) \cos^2 A}{\sin A(\cos A - \sin A)(\cos A + \sin A)} = \frac{\cos^2 A}{\sin A(\cos A + \sin A)} \)
Divide numerator and denominator by cos A:
\( = \frac{\cos A}{\sin A} \times \frac{1}{\frac{\cos A + \sin A}{\cos A}} = \frac{\cot A}{1 + \tan A} \)
In simple words: Replace sec² A with 1 + tan² A. Convert everything to sin and cos, then recognize and cancel (cos A - sin A).

Exam Tip: When (cos A - sin A) appears in both numerator and denominator, cancel it - but watch that you don't divide by zero if cos A = sin A.

Question 14(iv). \( \frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta} = 2 \sec^2 \theta \)
Answer: Combine the fractions on the left with a common denominator:
\( \frac{(1 - \sin \theta) + (1 + \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} = \frac{2}{1 - \sin^2 \theta} \)
Use \( 1 - \sin^2 \theta = \cos^2 \theta \):
\( = \frac{2}{\cos^2 \theta} = 2 \sec^2 \theta \)
In simple words: Find the common denominator (1 + sin θ)(1 - sin θ). This is a difference of squares that equals cos² θ.

Exam Tip: (1 + sin θ)(1 - sin θ) always equals 1 - sin² θ = cos² θ - this pattern shows up frequently in these proofs.

Question 15(i). tan²θ - sin²θ = tan²θ sin²θ
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta \]
\[ \Rightarrow \frac{\sin^2\theta - \sin^2\theta\cos^2\theta}{\cos^2\theta} \]
\[ \Rightarrow \frac{\sin^2\theta(1 - \cos^2\theta)}{\cos^2\theta} \]
\[ \Rightarrow \frac{\sin^2\theta \cdot \sin^2\theta}{\cos^2\theta} \]
\[ \Rightarrow \tan^2\theta \cdot \sin^2\theta \]

Since L.H.S. equals R.H.S., we have proven that tan²θ - sin²θ = tan²θ sin²θ.
In simple words: Take tan squared theta and subtract sin squared theta. Factor out sin squared from the numerator. Since 1 minus cos squared equals sin squared, you get tan squared times sin squared, which is the right side.

Exam Tip: Write each step clearly, showing the factorization of sin²θ(1 - cos²θ). Mark where you use the Pythagorean identity 1 - cos²θ = sin²θ — examiners look for this specific substitution.

Question 15(ii). \( \frac{\cos\theta}{1 - \tan\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta} = \cos\theta + \sin\theta \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{\cos\theta}{1 - \frac{\sin\theta}{\cos\theta}} - \frac{\sin^2\theta}{\cos\theta - \sin\theta} \]
\[ \Rightarrow \frac{\cos^2\theta}{\cos\theta - \sin\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta} \]
\[ \Rightarrow \frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta} \]
\[ \Rightarrow \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta - \sin\theta} \]
\[ \Rightarrow \cos\theta + \sin\theta \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{\cos\theta}{1 - \tan\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta} = \cos\theta + \sin\theta \).
In simple words: Rewrite the first fraction by replacing tan with sin over cos. Find a common denominator for both fractions. Factor the numerator as a difference of squares and cancel the common factor from numerator and denominator.

Exam Tip: Take care when handling the denominator 1 - tan θ - substitute tan as sin/cos first. The difference of squares factorization and cancellation of (cos θ - sin θ) are critical steps that show full understanding.

Question 16. Prove that: \( \frac{(1 + \sin\theta)^2 + (1 - \sin\theta)^2}{2\cos^2\theta} = \sec^2\theta + \tan^2\theta \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{(1 + \sin\theta)^2 + (1 - \sin\theta)^2}{2\cos^2\theta} \]
\[ \Rightarrow \frac{1 + \sin^2\theta + 2\sin\theta + 1 + \sin^2\theta - 2\sin\theta}{2\cos^2\theta} \]
\[ \Rightarrow \frac{2 + 2\sin^2\theta}{2\cos^2\theta} \]
\[ \Rightarrow \frac{2(1 + \sin^2\theta)}{2\cos^2\theta} \]
\[ \Rightarrow \frac{1 + \sin^2\theta}{\cos^2\theta} \]
\[ \Rightarrow \frac{1}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} \]
\[ \Rightarrow \sec^2\theta + \tan^2\theta \]

Since L.H.S. equals sec²θ + tan²θ = R.H.S., we have proven that \( \frac{(1 + \sin\theta)^2 + (1 - \sin\theta)^2}{2\cos^2\theta} = \sec^2\theta + \tan^2\theta \).
In simple words: Expand both squared terms in the numerator carefully - the cross terms cancel out. Combine like terms to get 2 plus 2 sin squared. Factor out 2, then divide the numerator by the denominator. Split the resulting fraction into two parts using sec and tan.

Exam Tip: When expanding (1 + sin θ)² and (1 - sin θ)², note that the 2 sin θ terms have opposite signs and cancel. After combining, you should have exactly 2 + 2sin²θ in the numerator - this is a common place where students make arithmetic mistakes.

Question 17(i). cosec⁴θ - cosec²θ = cot⁴θ + cot²θ
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \cosec^2\theta(\cosec^2\theta - 1) \]
\[ \Rightarrow \cosec^2\theta \cdot \cot^2\theta \]
\[ \Rightarrow (1 + \cot^2\theta) \cdot \cot^2\theta \]
\[ \Rightarrow \cot^2\theta + \cot^4\theta \]

Since L.H.S. equals R.H.S., we have proven that cosec⁴θ - cosec²θ = cot⁴θ + cot²θ.
In simple words: Factor out cosec squared theta from the left side. Use the identity that cosec squared minus 1 equals cot squared. Substitute the identity 1 plus cot squared equals cosec squared, then expand to match the right side.

Exam Tip: The key identity here is cosec²θ - 1 = cot²θ. Always show this step explicitly. Also clearly indicate when you substitute 1 + cot²θ for cosec²θ in the second line - examiners value seeing these substitutions written out.

Question 17(ii). 2sec²θ - sec⁴θ - 2cosec²θ + cosec⁴θ = cot⁴θ - tan⁴θ
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow 2(1 + \tan^2\theta) - (\sec^2\theta)^2 - 2(1 + \cot^2\theta) + (\cosec^2\theta)^2 \]
\[ \Rightarrow 2 + 2\tan^2\theta - (1 + \tan^2\theta)^2 - 2 - 2\cot^2\theta + (1 + \cot^2\theta)^2 \]
\[ \Rightarrow 2 + 2\tan^2\theta - (1 + \tan^4\theta + 2\tan^2\theta) - 2 - 2\cot^2\theta + (1 + \cot^4\theta + 2\cot^2\theta) \]
\[ \Rightarrow 2 + 2\tan^2\theta - 1 - \tan^4\theta - 2\tan^2\theta - 2 - 2\cot^2\theta + 1 + \cot^4\theta + 2\cot^2\theta \]
\[ \Rightarrow 2 - 2 + 2\tan^2\theta - 2\tan^2\theta + 2\cot^2\theta - 2\cot^2\theta + 1 - 1 + \cot^4\theta - \tan^4\theta \]
\[ \Rightarrow \cot^4\theta - \tan^4\theta \]

Since L.H.S. equals R.H.S., we have proven that 2sec²θ - sec⁴θ - 2cosec²θ + cosec⁴θ = cot⁴θ - tan⁴θ.
In simple words: Replace sec with 1 plus tan squared and cosec with 1 plus cot squared. Expand all the squared terms carefully. Most terms will cancel out - when you add and subtract like terms, you are left with cot to the fourth minus tan to the fourth.

Exam Tip: This proof requires careful bookkeeping of positive and negative signs. Work through the expansion of (1 + tan²θ)² and (1 + cot²θ)² step by step. At the end, verify that all intermediate terms genuinely cancel by listing which terms cancel with which.

Question 18(i). \( \frac{1 + \cos\theta - \sin^2\theta}{\sin\theta(1 + \cos\theta)} = \cot\theta \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{1 + \cos\theta - (1 - \cos^2\theta)}{\sin\theta(1 + \cos\theta)} \]
\[ \Rightarrow \frac{1 - 1 + \cos^2\theta + \cos\theta}{\sin\theta(1 + \cos\theta)} \]
\[ \Rightarrow \frac{\cos\theta(\cos\theta + 1)}{\sin\theta(1 + \cos\theta)} \]
\[ \Rightarrow \frac{\cos\theta}{\sin\theta} \]
\[ \Rightarrow \cot\theta \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{1 + \cos\theta - \sin^2\theta}{\sin\theta(1 + \cos\theta)} = \cot\theta \).
In simple words: Use sin squared equals 1 minus cos squared to rewrite the numerator. Simplify the constants that cancel. Factor out cos theta from what remains. Cancel the matching (1 plus cos theta) terms from top and bottom to get cos theta over sin theta.

Exam Tip: The critical identity is sin²θ = 1 - cos²θ. After substitution, simplify the 1's in the numerator first before factoring. Always show the cancellation of (1 + cos θ) as a separate step.

Question 18(ii). \( \frac{\tan^3\theta - 1}{\tan\theta - 1} = \sec^2\theta + \tan\theta \)
Answer: Use the factorization: a³ - b³ = (a - b)(a² + ab + b²)

Therefore: tan³θ - 1³ = (tan θ - 1)(tan²θ + tan θ + 1)

The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{(\tan\theta - 1)(\tan^2\theta + \tan\theta + 1)}{\tan\theta - 1} \]
\[ \Rightarrow \tan^2\theta + \tan\theta + 1 \]
\[ \Rightarrow \sec^2\theta - 1 + 1 + \tan\theta \]
\[ \Rightarrow \sec^2\theta + \tan\theta \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{\tan^3\theta - 1}{\tan\theta - 1} = \sec^2\theta + \tan\theta \).
In simple words: Factor the numerator as a difference of cubes. Cancel the matching (tan θ - 1) from numerator and denominator. Replace tan squared with sec squared minus 1 to finish the proof.

Exam Tip: Recognize the difference of cubes form immediately - this saves time. After cancellation, remember that tan²θ = sec²θ - 1, so tan²θ + tan θ + 1 becomes sec²θ + tan θ.

Question 19(i). \( \frac{1 + \cosec A}{\cosec A} = \frac{\cos^2 A}{1 - \sin A} \)
Answer:
Left side:
\[ \Rightarrow \frac{1 + \cosec A}{\cosec A} \]
\[ \Rightarrow \frac{1 + \frac{1}{\sin A}}{\frac{1}{\sin A}} \]
\[ \Rightarrow \frac{\frac{\sin A + 1}{\sin A}}{\frac{1}{\sin A}} \]
\[ \Rightarrow \frac{\sin A(\sin A + 1)}{\sin A} \]
\[ \Rightarrow \sin A + 1 \]

Right side:
\[ \Rightarrow \frac{1 - \sin^2 A}{1 - \sin A} \]
\[ \Rightarrow \frac{(1 - \sin A)(1 + \sin A)}{1 - \sin A} \]
\[ \Rightarrow 1 + \sin A \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{1 + \cosec A}{\cosec A} = \frac{\cos^2 A}{1 - \sin A} \).
In simple words: On the left, expand cosec as 1 over sin and simplify by dividing fractions. On the right, replace cos squared with 1 minus sin squared, then factor this as a difference of squares. Both sides simplify to 1 plus sin A.

Exam Tip: When working with cosec A in the denominator of a fraction, flip and multiply. On the right side, the substitution cos²A = 1 - sin²A is essential - mark this step clearly. The final factorization and cancellation should be shown explicitly.

Question 19(ii). \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \sqrt{\frac{(1 - \cos A)(1 + \cos A)}{(1 + \cos A)(1 + \cos A)}} \]
\[ \Rightarrow \sqrt{\frac{(1 - \cos^2 A)}{(1 + \cos A)^2}} \]
\[ \Rightarrow \sqrt{\frac{\sin^2 A}{(1 + \cos A)^2}} \]
\[ \Rightarrow \frac{\sin A}{1 + \cos A} \]

Since L.H.S. equals R.H.S., we have proven that \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A} \).
In simple words: Multiply the numerator and denominator inside the square root by (1 plus cos A). The numerator becomes a difference of squares: 1 minus cos squared, which equals sin squared. The denominator becomes (1 plus cos A) squared. When you take the square root, sin squared and the squared term both simplify.

Exam Tip: The key technique here is rationalizing by multiplying by (1 + cos A) on both numerator and denominator. This creates the perfect setup for using 1 - cos²A = sin²A. After taking the square root, be careful with signs - sin A is taken as positive.

Question 20(i). \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \tan A + \sec A \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \]
\[ \Rightarrow \sqrt{\frac{(1 + \sin A)^2}{(1 - \sin^2 A)}} \]
\[ \Rightarrow \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \]
\[ \Rightarrow \frac{1 + \sin A}{\cos A} \]
\[ \Rightarrow \frac{1}{\cos A} + \frac{\sin A}{\cos A} \]
\[ \Rightarrow \sec A + \tan A \]

Since L.H.S. equals R.H.S., we have proven that \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \tan A + \sec A \).
In simple words: Multiply the numerator and denominator inside the square root by (1 plus sin A). This makes the numerator a perfect square. The denominator becomes 1 minus sin squared, which is cos squared. Taking the square root gives you (1 plus sin A) divided by cos A. Split this into two fractions to get sec A plus tan A.

Exam Tip: Multiply both numerator and denominator by (1 + sin A) to create the perfect square form. Show clearly that 1 - sin²A = cos²A. When splitting the final fraction, you get sec A (which is 1/cos A) and tan A (which is sin A/cos A).

Question 20(ii). \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \cosec A - \cot A \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \sqrt{\frac{(1 - \cos A)(1 - \cos A)}{(1 + \cos A)(1 - \cos A)}} \]
\[ \Rightarrow \sqrt{\frac{(1 - \cos A)^2}{(1 - \cos^2 A)}} \]
\[ \Rightarrow \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}} \]
\[ \Rightarrow \frac{1 - \cos A}{\sin A} \]
\[ \Rightarrow \frac{1}{\sin A} - \frac{\cos A}{\sin A} \]
\[ \Rightarrow \cosec A - \cot A \]

Since L.H.S. equals R.H.S., we have proven that \( \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \cosec A - \cot A \).
In simple words: Multiply the numerator and denominator inside the square root by (1 minus cos A). This creates (1 minus cos A) squared in the numerator. The denominator becomes 1 minus cos squared, which equals sin squared. Taking the square root yields (1 minus cos A) over sin A. Split this into cosec A minus cot A.

Exam Tip: This proof mirrors the previous one but uses (1 - cos A) as the multiplier instead. After taking the square root, separating the fraction into 1/sin A and cos A/sin A gives you cosec A and cot A. Note the minus sign between them.

Question 21(i). \( \sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2\cosec A \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{\sqrt{(\sec A - 1)^2} + \sqrt{(\sec A + 1)^2}}{\sqrt{(\sec A + 1)(\sec A - 1)}} \]
\[ \Rightarrow \frac{\sec A - 1 + \sec A + 1}{\sqrt{\sec^2 A - 1}} \]
\[ \Rightarrow \frac{2\sec A}{\sqrt{\tan^2 A}} \]
\[ \Rightarrow \frac{2\sec A}{\tan A} \]
\[ \Rightarrow \frac{2}{\cos A} \cdot \frac{\cos A}{\sin A} \]
\[ \Rightarrow \frac{2}{\sin A} \]
\[ \Rightarrow 2\cosec A \]

Since L.H.S. equals R.H.S., we have proven that \( \sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2\cosec A \).
In simple words: Combine the two square root fractions by getting a common denominator. Simplify the numerator under the radical. Use the identity sec squared minus 1 equals tan squared. Divide sec A by tan A to get a result in terms of sin A, which equals 2 cosec A.

Exam Tip: Finding the common denominator is the first critical step. The numerator simplifies to 2 sec A because the 1's cancel. Remember that sec²A - 1 = tan²A, so √(tan²A) = tan A. Finally, when you divide sec A by tan A, simplify: (1/cos A) ÷ (sin A/cos A) = 1/sin A.

Question 21(ii). \( \frac{\cos A \cot A}{1 - \sin A} = 1 + \cosec A \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \cos A \times \frac{\cos A}{\sin A} \times \frac{1}{1 - \sin A} \]
\[ \Rightarrow \frac{\cos^2 A}{\sin A(1 - \sin A)} \]
\[ \Rightarrow \frac{1 - \sin^2 A}{\sin A(1 - \sin A)} \]
\[ \Rightarrow \frac{(1 - \sin A)(1 + \sin A)}{\sin A(1 - \sin A)} \]
\[ \Rightarrow \frac{1 + \sin A}{\sin A} \]
\[ \Rightarrow \frac{1}{\sin A} + \frac{\sin A}{\sin A} \]
\[ \Rightarrow \cosec A + 1 \]
\[ \Rightarrow 1 + \cosec A \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{\cos A \cot A}{1 - \sin A} = 1 + \cosec A \).
In simple words: Rewrite cot A as cos A over sin A. Replace cos squared with 1 minus sin squared. Factor this as a difference of squares. Cancel the matching (1 minus sin A) term from numerator and denominator. Split the final fraction into two parts to get 1 plus cosec A.

Exam Tip: After replacing cos²A with 1 - sin²A, the factorization (1 - sin A)(1 + sin A) should be immediate. Show the cancellation step clearly. The final step of splitting 1 + sin A over sin A into two fractions is straightforward but must be shown to receive full marks.

Question 22(i). \( \frac{1 + \tan A}{\sin A} + \frac{1 + \cot A}{\cos A} = 2(\sec A + \cosec A) \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{1 + \frac{\sin A}{\cos A}}{\sin A} + \frac{1 + \frac{\cos A}{\sin A}}{\cos A} \]
\[ \Rightarrow \frac{\cos A(1 + \frac{\sin A}{\cos A}) + \sin A(1 + \frac{\cos A}{\sin A})}{\sin A \cos A} \]
\[ \Rightarrow \frac{\cos A + \sin A + \sin A + \cos A}{\sin A \cos A} \]
\[ \Rightarrow \frac{2(\cos A + \sin A)}{\sin A \cos A} \]
\[ \Rightarrow \frac{2\cos A}{\sin A \cos A} + \frac{2\sin A}{\sin A \cos A} \]
\[ \Rightarrow \frac{2}{\sin A} + \frac{2}{\cos A} \]
\[ \Rightarrow 2\cosec A + 2\sec A \]
\[ \Rightarrow 2(\sec A + \cosec A) \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{1 + \tan A}{\sin A} + \frac{1 + \cot A}{\cos A} = 2(\sec A + \cosec A) \).
In simple words: Replace tan with sin over cos and cot with cos over sin. Find a common denominator for the two fractions. Expand and simplify the numerator carefully - you should get 2 times (cos A plus sin A). Split this fraction into two parts to get 2/sin A plus 2/cos A, which equals 2 times (cosec A plus sec A).

Exam Tip: When finding the common denominator sin A cos A, be careful to multiply each fraction's numerator correctly. After expanding (1 + sin A/cos A) and (1 + cos A/sin A), verify that you get cos A + sin A + sin A + cos A = 2(cos A + sin A). The final split into two fractions is key to reaching the answer form.

Question 22(ii). \( \frac{\sin A}{1 + \cot A} - \frac{\cos A}{1 + \tan A} = \sin A - \cos A \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{\sin A}{1 + \frac{\cos A}{\sin A}} - \frac{\cos A}{1 + \frac{\sin A}{\cos A}} \]
\[ \Rightarrow \frac{\sin A}{\left(\frac{\sin A + \cos A}{\sin A}\right)} - \frac{\cos A}{\left(\frac{\cos A + \sin A}{\cos A}\right)} \]
\[ \Rightarrow \frac{\sin A \times \sin A}{\sin A + \cos A} - \frac{\cos A \times \cos A}{\cos A + \sin A} \]
\[ \Rightarrow \frac{\sin^2 A - \cos^2 A}{\sin A + \cos A} \]
\[ \Rightarrow \frac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A + \cos A} \]
\[ \Rightarrow \sin A - \cos A \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{\sin A}{1 + \cot A} - \frac{\cos A}{1 + \tan A} = \sin A - \cos A \).
In simple words: Replace cot A and tan A with their fraction forms. Simplify each fraction's denominator by combining terms. Multiply and divide to get sin squared A and cos squared A with a common denominator. Factor the numerator as a difference of squares and cancel the common factor.

Exam Tip: After substituting cot A = cos A/sin A and tan A = sin A/cos A, simplify the denominators (1 + cot A) and (1 + tan A) to get single fractions. When you have sin²A and cos²A in the numerator, factor immediately as (sin A - cos A)(sin A + cos A). The cancellation with the denominator (sin A + cos A) gives the final answer.

Question 22(iii). sec⁴A - tan⁴A = 1 + 2tan²A
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow (\sec^2 A - \tan^2 A)(\sec^2 A + \tan^2 A) \]
\[ \Rightarrow 1 \times (\sec^2 A + \tan^2 A) \]
\[ \Rightarrow (\sec^2 A + \tan^2 A) \]
\[ \Rightarrow (1 + \tan^2 A + \tan^2 A) \]
\[ \Rightarrow 1 + 2\tan^2 A \]

Since L.H.S. equals R.H.S., we have proven that sec⁴A - tan⁴A = 1 + 2tan²A.
In simple words: Factor the left side as a difference of squares: (sec squared A minus tan squared A) times (sec squared A plus tan squared A). The first factor equals 1 by the Pythagorean identity. The second factor simplifies when you replace sec squared with 1 plus tan squared.

Exam Tip: Recognize immediately that this is a difference of squares pattern a⁴ - b⁴ = (a² - b²)(a² + b²). Use sec²A - tan²A = 1 explicitly. Then substitute sec²A = 1 + tan²A into the second factor to get 1 + tan²A + tan²A = 1 + 2tan²A.

Question 23(i). cosec⁶A - cot⁶A = 3cot²A cosec²A + 1
Answer: Use the factorization: a³ - b³ = (a - b)³ + 3ab(a - b)

The left side of the equation can be rewritten as follows:

\[ \Rightarrow \cosec^6 A - \cot^6 A = (\cosec^2 A - \cot^2 A)^3 + 3\cosec^2 A \cot^2 A(\cosec^2 A - \cot^2 A) \]
\[ \Rightarrow 1^3 + 3\cosec^2 A \cot^2 A \times 1 \]
\[ \Rightarrow 1 + 3\cosec^2 A \cot^2 A \]

Since L.H.S. equals R.H.S., we have proven that cosec⁶A - cot⁶A = 3cot²A cosec²A + 1.
In simple words: View cosec to the sixth and cot to the sixth as (cosec squared) cubed and (cot squared) cubed. Use the difference of cubes formula: a cubed minus b cubed. Since cosec squared minus cot squared equals 1, most terms disappear, leaving only 1 plus 3 cosec squared times cot squared.

Exam Tip: The formula a³ - b³ = (a - b)³ + 3ab(a - b) is less common than a³ - b³ = (a - b)(a² + ab + b²), but it is useful here. Apply it with a = cosec²A and b = cot²A. Remember that cosec²A - cot²A = 1, which simplifies the expression significantly.

Question 23(ii). sec⁶A - tan⁶A = 1 + 3tan²A + 3tan⁴A
Answer: Use the factorization: a³ - b³ = (a - b)³ + 3ab(a - b)

The left side of the equation can be rewritten as follows:

\[ \Rightarrow \sec^6 A - \tan^6 A = (\sec^2 A - \tan^2 A)^3 + 3\sec^2 A \tan^2 A(\sec^2 A - \tan^2 A) \]
\[ \Rightarrow 1^3 + 3\sec^2 A \tan^2 A \times 1 \]
\[ \Rightarrow 1 + 3\sec^2 A \tan^2 A \]
\[ \Rightarrow 1 + 3(1 + \tan^2 A)\tan^2 A \]
\[ \Rightarrow 1 + 3(\tan^4 A + \tan^2 A) \]
\[ \Rightarrow 1 + 3\tan^2 A + 3\tan^4 A \]

Since L.H.S. equals R.H.S., we have proven that sec⁶A - tan⁶A = 1 + 3tan²A + 3tan⁴A.
In simple words: Write sec to the sixth and tan to the sixth as perfect cubes. Apply the difference of cubes formula. Since sec squared minus tan squared equals 1, you get 1 plus 3 sec squared tan squared. Substitute sec squared equals 1 plus tan squared. Expand and collect like terms to get the final form.

Exam Tip: After using the difference of cubes formula, you have 1 + 3sec²A tan²A. The key step is substituting sec²A = 1 + tan²A into this expression. Expand carefully: 3(1 + tan²A)tan²A = 3tan²A + 3tan⁴A. Then add the initial 1 to get the final answer.

Question 24(i). \( \frac{\cot\theta + \cosec\theta - 1}{\cot\theta - \cosec\theta + 1} = \frac{1 + \cos\theta}{\sin\theta} \)
Answer: The left side of the equation can be rewritten as follows:

\[ \Rightarrow \frac{\frac{\cos\theta}{\sin\theta} + \frac{1}{\sin\theta} - 1}{\frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta} + 1} \]
\[ \Rightarrow \frac{\frac{\cos\theta + 1 - \sin\theta}{\sin\theta}}{\frac{\cos\theta - 1 + \sin\theta}{\sin\theta}} \]
\[ \Rightarrow \frac{\cos\theta + 1 - \sin\theta}{\cos\theta - 1 + \sin\theta} \]
\[ \Rightarrow \frac{\cos\theta - (1 - \sin\theta)}{\cos\theta + (1 - \sin\theta)} \]
\[ \Rightarrow \frac{\cos\theta - (1 - \sin\theta)}{\cos\theta + (1 - \sin\theta)} \times \frac{\cos\theta + (1 - \sin\theta)}{\cos\theta + (1 - \sin\theta)} \]
\[ \Rightarrow \frac{[\cos\theta - (1 - \sin\theta)]^2}{\cos^2\theta - (1 - \sin\theta)^2} \]
\[ \Rightarrow \frac{\cos^2\theta + (1 - \sin\theta)^2 - 2\cos\theta(1 - \sin\theta)}{\cos^2\theta - (1 + \sin^2\theta - 2\sin\theta)} \]
\[ \Rightarrow \frac{\cos^2\theta + \sin^2\theta + 1 + 2\cos\theta - 2\sin\theta - 2\sin\theta\cos\theta}{1 - 1 - \sin^2\theta + 2\sin\theta} \]
\[ \Rightarrow \frac{2 + 2\cos\theta - 2\sin\theta - 2\sin\theta\cos\theta}{2\sin\theta - \sin^2\theta} \]
\[ \Rightarrow \frac{2(1 + \cos\theta) - 2\sin\theta(1 + \cos\theta)}{2\sin\theta(1 - \sin\theta)} \]
\[ \Rightarrow \frac{2(1 + \cos\theta)(1 - \sin\theta)}{2\sin\theta(1 - \sin\theta)} \]
\[ \Rightarrow \frac{1 + \cos\theta}{\sin\theta} \]

Since L.H.S. equals R.H.S., we have proven that \( \frac{\cot\theta + \cosec\theta - 1}{\cot\theta - \cosec\theta + 1} = \frac{1 + \cos\theta}{\sin\theta} \).
In simple words: Replace cot and cosec with their definitions in terms of sin and cos. Simplify both numerator and denominator by combining fractions. Multiply both numerator and denominator by the conjugate to clear radicals. Expand carefully using difference of squares in the denominator. After simplification, factor out common terms in both parts of the fraction and cancel to get the right side.

Exam Tip: This is a complex proof with multiple algebraic steps. After substituting cot θ = cos θ/sin θ and cosec θ = 1/sin θ, simplify the complex fraction first. The rationalization step (multiplying by the conjugate) is crucial. Take time to expand both numerator and denominator correctly. Verify your final answer by substituting a test angle.

Question 24(ii). \( \frac{\sin\theta}{\cot\theta + \cosec\theta} = 2 + \frac{\sin\theta}{\cot\theta - \cosec\theta} \)
Answer:
Left side:
\[ \Rightarrow \frac{\sin\theta}{\frac{\cos\theta}{\sin\theta} + \frac{1}{\sin\theta}} \]
\[ \Rightarrow \frac{\sin^2\theta}{\cos\theta + 1} \]
\[ \Rightarrow \frac{1 - \cos^2\theta}{1 + \cos\theta} \]
\[ \Rightarrow \frac{(1 - \cos\theta)(1 + \cos\theta)}{1 + \cos\theta} \]
\[ \Rightarrow 1 - \cos\theta \]

Right side:
\[ \Rightarrow 2 + \frac{\sin\theta}{\frac{\cos\theta}{\sin\theta} - \frac{1}{\sin\theta}} \]
\[ \Rightarrow 2 + \frac{\sin^2\theta}{\cos\theta - 1} \]
\[ \Rightarrow \frac{2(\cos\theta - 1) + \sin^2\theta}{\cos\theta - 1} \]
\[ \Rightarrow \frac{2(\cos\theta - 1) + 1 - \cos^2\theta}{\cos\theta - 1} \]
\[ \Rightarrow \frac{2(\cos\theta - 1) + (1 + \cos\theta)(1 - \cos\theta)}{\cos\theta - 1} \]
\[ \Rightarrow \frac{2(\cos\theta - 1) - (\cos\theta - 1)(1 + \cos\theta)}{\cos\theta - 1} \]
\[ \Rightarrow \frac{(\cos\theta - 1)[2 - (1 + \cos\theta)]}{\cos\theta - 1} \]
\[ \Rightarrow 2 - 1 - \cos\theta \]
\[ \Rightarrow 1 - \cos\theta \]

Since L.H.S. = 1 - cos θ = R.H.S., we have proven that \( \frac{\sin\theta}{\cot\theta + \cosec\theta} = 2 + \frac{\sin\theta}{\cot\theta - \cosec\theta} \).
In simple words: On the left, replace cot and cosec, then simplify to get sin squared over (1 plus cos). Use sin squared equals 1 minus cos squared to factor and cancel. On the right, work with the 2 plus the fraction separately. Get a common denominator and substitute sin squared. Factor and simplify to match the left side result of 1 minus cos theta.

Exam Tip: Both sides must equal 1 - cos θ. For the left side, the key step is replacing sin²θ with 1 - cos²θ and factoring. For the right side, carefully manage the 2 in front - convert it to a fraction with denominator (cos θ - 1). Use sin²θ = 1 - cos²θ = (1 + cos θ)(1 - cos θ), and note that (1 - cos θ) = -(cos θ - 1).

Question 25(i). Prove that \( (\sin \theta + \cos \theta)(\sec \theta + \cosec \theta) = 2 + \sec \theta \cosec \theta \)
Answer: The left side of the given equation can be expressed as follows:
\( \implies (\sin \theta + \cos \theta)\left(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\right) \)
\( \implies \frac{(\sin \theta + \cos \theta)(\sin \theta + \cos \theta)}{\sin \theta \cos \theta} \)
\( \implies \frac{\sin^2\theta + \cos^2\theta + 2 \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( \implies \frac{1 + 2 \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( \implies \frac{1}{\sin \theta \cos \theta} + \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( \implies \cosec \theta \sec \theta + 2 \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Rewrite the secant and cosecant using their reciprocal definitions, combine the fractions inside the second bracket, and use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to get the result.

Exam Tip: Remember to recognize that \( \sin^2 \theta + \cos^2 \theta = 1 \) is the key step. Always convert secant and cosecant to their reciprocal forms at the start.

Question 25(ii). Prove that \( (\cosec A - \sin A)(\sec A - \cos A)\sec^2 A = \tan A \)
Answer: The left side of the given equation can be expressed as:
\( \implies \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) \times \frac{1}{\cos^2 A} \)
\( \implies \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) \times \frac{1}{\cos^2 A} \)
Since \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \):
\( \implies \frac{\cos^2 A \sin^2 A}{\sin A \cos A} \times \frac{1}{\cos^2 A} \)
Dividing numerator and denominator by \( \sin A \cos A \):
\( \implies \frac{\sin A \cos A}{\cos^2 A} \)
\( \implies \frac{\sin A}{\cos A} \)
\( \implies \tan A \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Express cosecant and secant as reciprocals, use the Pythagorean identity to simplify the numerators, and then cancel common terms to arrive at tangent.

Exam Tip: The Pythagorean identities \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \) are essential here. Do not rush the cancellation step.

Question 26(i). Prove that \( \frac{\sin^3 A + \cos^3 A}{\sin A + \cos A} + \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 2 \)
Answer: We use the algebraic identities:
\( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
and
\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
The left side becomes:
\( \implies \frac{(\sin A + \cos A)(\sin^2 A - \sin A \cos A + \cos^2 A)}{\sin A + \cos A} + \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A - \cos A} \)
\( \implies (\sin^2 A - \sin A \cos A + \cos^2 A) + (\sin^2 A + \sin A \cos A + \cos^2 A) \)
\( \implies \sin^2 A + \cos^2 A - \sin A \cos A + \sin A \cos A + \sin^2 A + \cos^2 A \)
\( \implies 1 + 1 = 2 \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Factor the sum and difference of cubes in each fraction, cancel the common factors from numerator and denominator, and use \( \sin^2 A + \cos^2 A = 1 \) to finish.

Exam Tip: Always factor cube expressions using the standard formulas. The cancellation of numerator and denominator is straightforward once factorization is done correctly.

Question 26(ii). Prove that \( \frac{\tan^2 A}{1 + \tan^2 A} + \frac{\cot^2 A}{1 + \cot^2 A} = 1 \)
Answer: The left side of the equation can be expressed as:
\( \implies \frac{\tan^2 A}{1 + \tan^2 A} + \frac{\frac{1}{\tan^2 A}}{1 + \frac{1}{\tan^2 A}} \)
\( = \frac{\tan^2 A}{1 + \tan^2 A} + \frac{\frac{1}{\tan^2 A}}{\frac{\tan^2 A + 1}{\tan^2 A}} \)
\( = \frac{\tan^2 A}{1 + \tan^2 A} + \frac{1}{\tan^2 A + 1} \)
\( = \frac{\tan^2 A + 1}{1 + \tan^2 A} = 1 \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Replace cotangent with the reciprocal of tangent, simplify the complex fraction by multiplying by the reciprocal, and combine the terms to get the final answer.

Exam Tip: When working with cotangent, immediately convert it to a reciprocal of tangent. This substitution makes the algebraic simplification much easier to manage.

Question 27(i). Prove that \( \frac{1}{\sec A + \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A - \tan A} \)
Answer: The equation can be rewritten as:
\( \frac{1}{\sec A + \tan A} + \frac{1}{\sec A - \tan A} = \frac{2}{\cos A} \)
The left side becomes:
\( \implies \frac{\sec A - \tan A + \sec A + \tan A}{(\sec A - \tan A)(\sec A + \tan A)} \)
\( = \frac{2 \sec A}{\sec^2 A - \tan^2 A} \)
\( = \frac{2 \sec A}{1} = 2 \sec A \)
\( = \frac{2}{\cos A} \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Add the two fractions on the left by finding a common denominator, apply the difference of squares to the denominator, and use the identity \( \sec^2 A - \tan^2 A = 1 \) to simplify.

Exam Tip: Recognize that the denominator becomes a difference of squares, which simplifies nicely using the trigonometric identity. This is a key insight for solving this problem quickly.

Question 27(ii). Prove that \( (\sin A + \sec A)^2 + (\cos A + \cosec A)^2 = (1 + \sec A \cosec A)^2 \)
Answer: The left side becomes:
\( \implies \sin^2 A + \sec^2 A + 2 \sin A \sec A + \cos^2 A + \cosec^2 A + 2 \cos A \cosec A \)
\( = \sin^2 A + \cos^2 A + \sec^2 A + \cosec^2 A + 2 \sin A \sec A + 2 \cos A \cosec A \)
\( = 1 + \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} + \frac{2 \sin A}{\cos A} + \frac{2 \cos A}{\sin A} \)
\( = 1 + \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} + \frac{2 \sin^2 A + 2 \cos^2 A}{\sin A \cos A} \)
\( = 1 + \frac{1}{\sin^2 A \cos^2 A} + \frac{2(\sin^2 A + \cos^2 A)}{\sin A \cos A} \)
\( = 1 + \frac{1}{\sin A \cos A} + \frac{2}{\sin A \cos A} \)
\( = \left(1 + \frac{1}{\sin A \cos A}\right)^2 \)
\( = (1 + \sec A \cosec A)^2 \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Expand both squared terms, group the sine and cosine squared terms together to equal 1, and then factor the result as a perfect square.

Exam Tip: Organize your work carefully by grouping like terms. The key is recognizing that \( \sin^2 A + \cos^2 A = 1 \) appears multiple times and that the final expression is a perfect square.

Question 27(iii). Prove that \( \frac{\tan A + \sin A}{\tan A - \sin A} = \frac{\sec A + 1}{\sec A - 1} \)
Answer: The left side becomes:
\( \implies \frac{\frac{\sin A}{\cos A} + \sin A}{\frac{\sin A}{\cos A} - \sin A} \)
\( = \frac{\sin A\left(\frac{1}{\cos A} + 1\right)}{\sin A\left(\frac{1}{\cos A} - 1\right)} \)
\( = \frac{\frac{1}{\cos A} + 1}{\frac{1}{\cos A} - 1} \)
\( = \frac{\sec A + 1}{\sec A - 1} \)
Since L.H.S. = R.H.S., the statement is proved.
In simple words: Express tangent as sine divided by cosine, factor out the sine from both numerator and denominator, and simplify to get secant in both.

Exam Tip: Always convert tangent to the sine and cosine ratio when dealing with mixed trigonometric functions. Factoring common terms makes cancellation straightforward.

Question 28. If \( \sin \theta + \cos \theta = \sqrt{2} \sin(90° - \theta) \), show that \( \cot \theta = \sqrt{2} + 1 \)
Answer: Given:
\( \sin \theta + \cos \theta = \sqrt{2} \sin(90° - \theta) \)
\( \implies \sin \theta + \cos \theta = \sqrt{2} \cos \theta \)
Dividing both sides by \( \sin \theta \):
\( \implies 1 + \cot \theta = \sqrt{2} \cot \theta \)
\( \implies 1 = \sqrt{2} \cot \theta - \cot \theta \)
\( \implies 1 = \cot \theta(\sqrt{2} - 1) \)
\( \implies \cot \theta = \frac{1}{\sqrt{2} - 1} \)
Rationalizing:
\( \implies \cot \theta = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \)
\( \implies \cot \theta = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \)
Hence, the statement is proved.
In simple words: Use the complementary angle identity to simplify the right side, divide by sine, collect the cotangent terms, and rationalize the denominator by multiplying by the conjugate.

Exam Tip: Rationalizing the denominator is crucial in this problem. Always multiply by the conjugate when you have a square root in the denominator.

Question 29. If \( 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \), where \( 0° \leq \theta \leq 90° \), find the value of \( \theta \)
Answer: Given:
\( 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \)
\( \implies 4 \sin^2 \theta + 3 \sin^2 \theta + 3 \cos^2 \theta = 4 \)
\( \implies 4 \sin^2 \theta + 3(\sin^2 \theta + \cos^2 \theta) = 4 \)
\( \implies 4 \sin^2 \theta + 3 = 4 \)
\( \implies 4 \sin^2 \theta = 1 \)
\( \implies \sin^2 \theta = \frac{1}{4} \)
Taking the square root of both sides:
\( \implies \sin \theta = \frac{1}{2} \)
\( \implies \theta = 30° \)
Hence, the value of \( \theta = 30° \).
In simple words: Split the equation to use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), isolate the sine squared term, and take the square root to find the angle.

Exam Tip: Breaking up the coefficient to make the identity visible is a common technique. Remember that \( \sin 30° = \frac{1}{2} \) is a standard value you should recognize immediately.

Question 30. If \( \sec \theta + \tan \theta = m \) and \( \sec \theta - \tan \theta = n \), prove that \( mn = 1 \)
Answer:
\( mn = (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) \)
\( mn = (\sec^2 \theta - \tan^2 \theta) \)
Using the trigonometric identity \( \sec^2 \theta - \tan^2 \theta = 1 \):
\( \therefore mn = 1 \)
Hence, the statement is proved.
In simple words: Multiply the two given expressions together and recognize that the result is a difference of squares, which equals 1 by the Pythagorean identity.

Exam Tip: This is a direct application of the difference of squares formula and the Pythagorean identity. Recognizing when to use these identities speeds up the solution significantly.

Question 31. If \( x = a \sec \theta + b \tan \theta \) and \( y = a \tan \theta + b \sec \theta \), prove that \( x^2 - y^2 = a^2 - b^2 \)
Answer: Given:
\( x = a \sec \theta + b \tan \theta \) and \( y = a \tan \theta + b \sec \theta \)
\( \therefore x^2 - y^2 = (a \sec \theta + b \tan \theta)^2 - (a \tan \theta + b \sec \theta)^2 \)
\( \implies x^2 - y^2 = a^2 \sec^2 \theta + b^2 \tan^2 \theta + 2ab \sec \theta \tan \theta - (a^2 \tan^2 \theta + b^2 \sec^2 \theta + 2ab \sec \theta \tan \theta) \)
\( \implies x^2 - y^2 = a^2 \sec^2 \theta + b^2 \tan^2 \theta + 2ab \sec \theta \tan \theta - a^2 \tan^2 \theta - b^2 \sec^2 \theta - 2ab \sec \theta \tan \theta \)
\( \implies x^2 - y^2 = a^2(\sec^2 \theta - \tan^2 \theta) - b^2(\sec^2 \theta - \tan^2 \theta) \)
\( \implies x^2 - y^2 = a^2 - b^2 \)
Hence, the statement is proved.
In simple words: Expand both squared terms, group the secant and tangent squared differences, and apply the identity \( \sec^2 \theta - \tan^2 \theta = 1 \) to factor and simplify.

Exam Tip: After expanding both squares, group terms carefully. The key is recognizing that \( \sec^2 \theta - \tan^2 \theta \) appears as a common factor in both the \( a \) and \( b \) terms.

Question 32. If \( x = h + a \cos \theta \) and \( y = k + a \sin \theta \), prove that \( (x - h)^2 + (y - k)^2 = a^2 \)
Answer: Given:
\( x = h + a \cos \theta \) and \( y = k + a \sin \theta \)
\( \therefore (x - h)^2 + (y - k)^2 = (h + a \cos \theta - h)^2 + (k + a \sin \theta - k)^2 \)
\( \implies (x - h)^2 + (y - k)^2 = (a \cos \theta)^2 + (a \sin \theta)^2 \)
\( \implies (x - h)^2 + (y - k)^2 = a^2 \cos^2 \theta + a^2 \sin^2 \theta \)
\( \implies (x - h)^2 + (y - k)^2 = a^2(\cos^2 \theta + \sin^2 \theta) \)
\( \implies (x - h)^2 + (y - k)^2 = a^2 \)
Hence, the statement is proved.
In simple words: Cancel the constant terms to isolate the trigonometric parts, factor out the common \( a \), and apply the fundamental identity \( \cos^2 \theta + \sin^2 \theta = 1 \).

Exam Tip: This proof shows that the equation describes a circle with center \( (h, k) \) and radius \( a \). Understanding the geometric meaning helps solidify the algebraic steps.

Multiple Choice Questions

Question 1. \( \cot^2 \theta - \frac{1}{\sin^2 \theta} \) is equal to
(a) 1
(b) - 1
(c) \( \sin^2 \theta \)
(d) \( \sec^2 \theta \)
Answer: (b) - 1
In simple words: Convert cotangent to cosine over sine, find a common denominator, use the Pythagorean identity to simplify the numerator, and divide to get the final result.

Exam Tip: When you see a difference of trigonometric expressions, always find a common denominator. This reveals the underlying Pythagorean identity clearly.

Question 2. \( (\sec^2 \theta - 1)(1 - \cosec^2 \theta) \) is equal to
(a) - 1
(b) 1
(c) 0
(d) 2
Answer: (a) - 1
In simple words: Use the Pythagorean identities to replace \( \sec^2 \theta - 1 \) with \( \tan^2 \theta \) and \( 1 - \cosec^2 \theta \) with \( -\cot^2 \theta \). Multiply to get the answer.

Exam Tip: Remember that \( \sec^2 \theta = 1 + \tan^2 \theta \) and \( \cosec^2 \theta = 1 + \cot^2 \theta \). These rearrangements are essential for solving such problems quickly.

Question 3. \( \frac{\tan^2 \theta}{1 + \tan^2 \theta} \) is equal to
(a) \( 2\sin^2 \theta \)
(b) \( 2\cos^2 \theta \)
(c) \( \sin^2 \theta \)
(d) \( \cos^2 \theta \)
Answer: (c) \( \sin^2 \theta \)
In simple words: Replace \( 1 + \tan^2 \theta \) with \( \sec^2 \theta \), write tangent and secant in terms of sine and cosine, and simplify the complex fraction.

Exam Tip: Knowing that \( 1 + \tan^2 \theta = \sec^2 \theta \) immediately simplifies this fraction. Always look for Pythagorean identities in the denominator.

Question 4. \( (\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 \) is equal to
(a) - 2
(b) 0
(c) 1
(d) 2
Answer: (d) 2
In simple words: Expand both squared binomials, the cross terms cancel, and the sum simplifies to twice the sum of cosine squared and sine squared.

Exam Tip: When two squared expressions have opposite middle terms, they often simplify nicely when added together. Watch for such cancellations.

Question 5. \( (\sec A + \tan A)(1 - \sin A) \) is equal to
(a) \( \sec A \)
(b) \( \sin A \)
(c) \( \cosec A \)
(d) \( \cos A \)
Answer: (d) \( \cos A \)
In simple words: Write secant and tangent as reciprocals and ratios of sine and cosine. Combine the terms in the first bracket, use the difference of squares formula, and simplify.

Exam Tip: Look for the pattern \( (1 + \sin A)(1 - \sin A) = 1 - \sin^2 A = \cos^2 A \) when working with these expressions.

Question 6. \( \frac{1 + \cot^2 A}{1 + \tan^2 A} \) is equal to
(a) \( \sec^2 A \)
(b) - 1
(c) \( \cot^2 A \)
(d) \( \tan^2 A \)
Answer: (d) \( \tan^2 A \)
In simple words: Replace \( 1 + \cot^2 A \) with \( \cosec^2 A \) and \( 1 + \tan^2 A \) with \( \sec^2 A \). Express these in terms of sine and cosine, then simplify to get tangent squared.

Exam Tip: Always substitute the Pythagorean identities \( 1 + \tan^2 A = \sec^2 A \) and \( 1 + \cot^2 A = \cosec^2 A \) first for easy simplification.

Question 7. If \( \sec \theta - \tan \theta = k \), then the value of \( \sec \theta + \tan \theta \) is
(a) \( 1 - \frac{1}{k} \)
(b) \( 1 - k \)
(c) \( 1 + k \)
(d) \( \frac{1}{k} \)
Answer: (d) \( \frac{1}{k} \)
In simple words: Use the Pythagorean identity \( \sec^2 \theta - \tan^2 \theta = 1 \), which factors as a difference of squares. Set up an equation using the given condition and solve for the unknown expression.

Exam Tip: This problem uses the fact that the product of these two expressions equals 1. Recognizing this relationship allows you to find one in terms of the other without extensive calculation.

Question 8. If \( \theta \) is an acute angle of a right triangle, then the value of \( \sin \theta \cos(90° - \theta) + \cos \theta \sin(90° - \theta) \) is
(a) 0
(b) \( 2 \sin \theta \cos \theta \)
(c) 1
(d) \( 2 \sin^2 \theta \)
Answer: (c) 1
In simple words: Apply the complementary angle formulas - cosine of \( 90° - \theta \) equals sine of \( \theta \), and sine of \( 90° - \theta \) equals cosine of \( \theta \). The expression becomes sine squared plus cosine squared, which equals 1.

Exam Tip: Complementary angle identities are essential for these types of questions. Memorize \( \cos(90° - \theta) = \sin \theta \) and \( \sin(90° - \theta) = \cos \theta \).

Question 9. The value of \( \cos 65° \sin 25° + \sin 65° \cos 25° \) is
(a) 0
(b) 1
(c) 2
(d) 4
Answer: (b) 1
In simple words: Notice that \( 65° = 90° - 25° \), so use the complementary angle formulas. The expression becomes sine of \( 25° \) times sine of \( 25° \) plus cosine of \( 25° \) times cosine of \( 25° \), which equals 1.

Exam Tip: Always look for complementary angles (angles that add to 90 degrees). This approach transforms the problem into a simpler form using basic identities.

Question 9. Using the values, work out cos 65° sin 25° + sin 65° cos 25°.
Answer: We can rewrite the expression by noting that sin 25° = sin(90° - 65°) = cos 65° and cos 25° = cos(90° - 65°) = sin 65°. Substituting these into the original expression gives us cos 65° cos 65° + sin 65° sin 65°, which simplifies to cos² 65° + sin² 65°. Using the Pythagorean identity, this equals 1.
In simple words: When you add the squares of the sine and cosine of any angle, you always get 1. That is a key rule in trigonometry.

Exam Tip: Look for complementary angle relationships (angles that add to 90°) — they let you replace one trig ratio with another, which often simplifies the problem.

Question 10. The value of 3 tan² 26° - 3 cosec² 64° is
(a) 0
(b) 3
(c) -3
(d) -1
Answer: (c) -3
In simple words: First, use the fact that cosec² 64° can be written in terms of 26° (since 64° and 26° add to 90°). Then apply the identity that links sec² and tan², which gives you -1. Multiply by 3 to get -3.

Exam Tip: Always check if angles add to 90° — complementary angles let you swap between trig functions and often reveal hidden identities.

Question 11. Statement (i): sin² θ + cos² θ = 1. Statement (ii): cosec² θ + cot² θ = 1. Which of the following is valid?
(a) only (i)
(b) only (ii)
(c) both (i) and (ii)
(d) neither (i) nor (ii)
Answer: (a) only (i)
In simple words: The first statement is a fundamental identity that holds for all angles. The second statement is wrong - the correct identity is cosec² θ - cot² θ = 1, not the sum shown.

Exam Tip: Memorise the three Pythagorean identities correctly: sin² θ + cos² θ = 1, sec² θ - tan² θ = 1, and cosec² θ - cot² θ = 1. Common mistakes happen when students confuse plus and minus signs.

Question. Assertion (A): sec² 23° - tan² 23° = 1. Reason (R): cos 60° = 1/2
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is an incorrect reason for Assertion (A).
Answer: (a) Assertion (A) is true, Reason (R) is false.
In simple words: The assertion follows from the identity sec² θ - tan² θ = 1, which holds for any angle including 23°. The reason states a different trig fact that has nothing to do with why the assertion is true. Also, cos 60° = 1/2 is correct, but it does not explain or prove the assertion.

Exam Tip: In assertion-reason questions, always check two things: whether each statement is true on its own, and whether the reason actually explains why the assertion is true. A true reason that is unrelated to the assertion means you choose "both correct, but wrong reason."

Question. Assertion (A): For 0 < θ ≤ 90°, cosec θ - cot θ and cosec θ + cot θ are reciprocals of each other. Reason (R): cosec² θ - cot² θ = 1
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is an incorrect reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The identity cosec² θ - cot² θ = 1 can be factored as (cosec θ - cot θ)(cosec θ + cot θ) = 1. This factorisation shows directly that the two expressions multiply to 1, which is exactly what it means for them to be reciprocals. So the reason fully explains the assertion.

Exam Tip: When you see a difference of squares (a² - b²), try to factor it. Factoring often reveals a key relationship that shows why an assertion is true.

Question. Assertion (A): cosec² 54° - cot² 54° = 1. Reason (R): cosec² θ - cot² θ = 1 for all values of θ, 0° < θ ≤ 90°.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is an incorrect reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: The reason is a universal identity - it is true for every angle in the given range. Since 54° falls within that range, the assertion automatically follows from the reason. This is a straightforward application of a general rule to a specific angle.

Exam Tip: Whenever a reason states a general identity and the assertion applies that same identity to a specific angle within the valid range, the reason is always the correct explanation.

Question. Assertion (A): 1 + sec² θ = tan² θ is a trigonometric identity. Reason (R): An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of the angles involved.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is an incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true.
In simple words: The correct identity is 1 + tan² θ = sec² θ, not what the assertion says. The reason correctly defines what a trigonometric identity is, but the assertion does not meet that definition because it is not true for all angles.

Exam Tip: Memorise the three key Pythagorean identities and their exact forms. A single sign error (plus instead of minus, or vice versa) makes an identity false - watch for this trap in assertion-type questions.

Question 1(i). If θ is an acute angle and cosec θ = √5, find the value of cot θ - cos θ.
Answer: Since cosec θ = √5, we have sin θ = 1/√5. Using the Pythagorean identity, cos² θ = 1 - sin² θ = 1 - 1/5 = 4/5, so cos θ = 2/√5. Then cot θ = cos θ / sin θ = (2/√5) / (1/√5) = 2. Therefore, cot θ - cos θ = 2 - 2/√5 = 2(1 - 1/√5) = 2(√5 - 1)/√5.
In simple words: Find sin θ from cosec θ. Then use sin² θ + cos² θ = 1 to get cos θ. Finally, divide cos by sin to find cot, and subtract.

Exam Tip: Always rationalise your final answer if it contains a square root in the denominator. Show each step clearly - examiners award marks for method, not just the final number.

Question 1(ii). If θ is an acute angle and tan θ = 8/15, find the value of sec θ + cosec θ.
Answer: From tan θ = 8/15, we calculate sec² θ = 1 + tan² θ = 1 + 64/225 = 289/225, so sec θ = 17/15. Similarly, cot θ = 15/8, and cosec² θ = 1 + cot² θ = 1 + 225/64 = 289/64, so cosec θ = 17/8. Adding them together: sec θ + cosec θ = 17/15 + 17/8 = (17 × 8 + 17 × 15)/120 = (136 + 255)/120 = 391/120 = 3 31/120.
In simple words: Use the identities sec² θ = 1 + tan² θ and cosec² θ = 1 + cot² θ to find both values. Then add them and simplify the fraction.

Exam Tip: When adding fractions, find a common denominator - here, 120 = 15 × 8. Express the final answer as both an improper fraction and a mixed number for full marks.

Question 2(i). Evaluate 2 × (cos² 20° + cos² 70°) / (sin² 25° + sin² 65°) - tan 45° + tan 13° / (tan 23° tan 30° tan 67° tan 77°).
Answer: Apply complementary angle identities: cos 70° = sin 20°, sin 65° = cos 25°, tan 67° = cot 23°, and tan 77° = cot 13°. The first fraction becomes 2(cos² 20° + sin² 20°) / (sin² 25° + cos² 25°) = 2 × 1 / 1 = 2. For the second part, tan 45° = 1, and the denominator simplifies: tan 23° tan 30° cot 23° cot 13° = tan 30° cot 13° tan 13° = tan 30° = 1/√3. The numerator -1 + 1 = 0, so this term contributes 0/1 = 0. The last term is tan 13° cot 13° tan 23° cot 23° tan 30° = 1 × 1 × (1/√3) = 1/√3. Combining all parts: 2 - 1 + 1/√3 = 1 + 1/√3 = (√3 + 1)/√3 = (√3 + 1)√3 / 3 = (3 + √3)/√3.
In simple words: Break the expression into parts. Use complementary angles to pair up trig ratios. Many products will cancel to 1. Simplify step by step.

Exam Tip: Look for complementary angles (pairs that add to 90°) and reciprocal pairs (like tan and cot of the same angle). These often cancel or simplify quickly.

Question 2(ii). Evaluate (sin² 25° + sin² 65°) / (cos² 20° + cos² 70°) + sin² 63° + cos 63° sin 27°.
Answer: Using complementary angles, sin 65° = cos 25° and cos 70° = sin 20°. The first fraction becomes (sin² 25° + cos² 25°) / (cos² 20° + sin² 20°) = 1/1 = 1. For the second part, sin 27° = sin(90° - 63°) = cos 63°, so we have sin² 63° + cos 63° cos 63° = sin² 63° + cos² 63° = 1. Adding both parts: 1 + 1 = 2.
In simple words: Replace each angle with its complementary form. Watch for pairs like sin² + cos² that always equal 1. Add the results together.

Exam Tip: Whenever you see sin² and cos² of angles that add to 90° in the same expression, they will simplify to 1. This saves time on lengthy calculations.

Question 3. If (4/3)(sec² 59° - cot² 31°) - (2/3) sin 90° + 3 tan² 56° tan² 34° = x/3, then find the value of x.
Answer: Rewrite using complementary angles: cot 31° = cot(90° - 59°) = tan 59°, and tan 34° = tan(90° - 56°) = cot 56°. The expression becomes (4/3)(sec² 59° - tan² 59°) - (2/3)(1) + 3 tan² 56° cot² 56°. Using the Pythagorean identity sec² 59° - tan² 59° = 1, and the fact that tan 56° cot 56° = 1, this simplifies to (4/3)(1) - 2/3 + 3(1) = 4/3 - 2/3 + 3 = 2/3 + 3 = 11/3. Therefore x = 11.
In simple words: Use complementary angle identities and Pythagorean identities to replace complex expressions with 1. Simplify arithmetic step by step, keeping fractions organised.

Exam Tip: Always look for reciprocal products (tan × cot = 1) and Pythagorean identities that simplify to 1. They reduce a complex-looking problem to simple arithmetic.

Question 4(i). Prove the following identity: cos A / (1 - sin A) + cos A / (1 + sin A) = 2 sec A.
Answer: Combine the two fractions over a common denominator: [cos A(1 + sin A) + cos A(1 - sin A)] / [(1 - sin A)(1 + sin A)]. The numerator becomes cos A + cos A sin A + cos A - cos A sin A = 2 cos A. The denominator simplifies to 1 - sin² A = cos² A. Therefore, the left side equals 2 cos A / cos² A = 2 / cos A = 2 sec A, which is the right side. Hence proved.
In simple words: Add the two fractions by finding a common denominator. Cancel terms in the numerator. Use the identity sin² + cos² = 1 to simplify the denominator. The result is 2 sec A.

Exam Tip: When proving identities, always combine fractions on the more complex side first. Look for identities like sin² + cos² = 1 that will help simplify the final result.

Question 4(ii). Prove the following identity: cos A / (cosec A + 1) + cos A / (cosec A - 1) = 2 tan A.
Answer: Combine the fractions: [cos A(cosec A - 1) + cos A(cosec A + 1)] / [(cosec A + 1)(cosec A - 1)]. The numerator simplifies to cos A cosec A - cos A + cos A cosec A + cos A = 2 cos A cosec A. The denominator is cosec² A - 1 = cot² A. Thus, the left side becomes 2 cos A cosec A / cot² A = 2 cos A × (1 / sin A) / (cot² A) = 2 cot A / cot² A = 2 / cot A = 2 tan A. Hence proved.
In simple words: Add the fractions using a common denominator. Simplify the numerator and use the identity cosec² A - 1 = cot² A. The final ratio simplifies to 2 tan A.

Exam Tip: When dealing with cosec and cot, use the identity cosec² θ - cot² θ = 1 to simplify denominators quickly.

Question 4(iii). Prove the following identity: (cos θ - sin θ)(1 + tan θ) / (2 cos² θ - 1) = sec θ.
Answer: Expand the numerator: (cos θ - sin θ)(1 + sin θ / cos θ) = (cos θ - sin θ)(cos θ + sin θ) / cos θ = (cos² θ - sin² θ) / cos θ. The denominator 2 cos² θ - 1 can be rewritten as 2 cos² θ - (sin² θ + cos² θ) = cos² θ - sin² θ. Therefore, the left side becomes (cos² θ - sin² θ) / [cos θ(cos² θ - sin² θ)] = 1 / cos θ = sec θ. Hence proved.
In simple words: Expand (1 + tan θ) as (cos θ + sin θ) / cos θ. Use the difference of squares formula on the numerator. The (cos² θ - sin² θ) in numerator and denominator cancel. What remains is 1 / cos θ = sec θ.

Exam Tip: Spot the difference of squares formula (a² - b²) - it often appears in both numerator and denominator and will cancel to simplify the proof.

Question 5(i). Prove the following identity: sin² θ + cos⁴ θ = cos² θ + sin⁴ θ.
Answer: Start with the left side: sin² θ + cos⁴ θ. Substitute sin² θ = 1 - cos² θ and cos⁴ θ = (cos² θ)² = (1 - sin² θ)². Expanding (1 - sin² θ)² gives 1 + sin⁴ θ - 2 sin² θ. So the left side becomes (1 - cos² θ) + (1 + sin⁴ θ - 2 sin² θ) = 2 - cos² θ + sin⁴ θ - 2(1 - cos² θ) = 2 - cos² θ + sin⁴ θ - 2 + 2 cos² θ = cos² θ + sin⁴ θ, which is the right side. Hence proved.
In simple words: Substitute sin² θ = 1 - cos² θ to rewrite everything in terms of cos. Expand and simplify carefully. The left and right sides become identical.

Exam Tip: When an identity has both sin and cos on both sides, try converting everything to one variable using sin² θ + cos² θ = 1. Expand squared terms carefully and combine like terms.

Question 5(ii). Prove the following identity: cot θ / (cosec θ + 1) + (cosec θ + 1) / cot θ = 2 sec θ.
Answer: Combine the left side over a common denominator: [cot² θ + (cosec θ + 1)²] / [cot θ(cosec θ + 1)]. Expand the numerator: cot² θ + cosec² θ + 1 + 2 cosec θ. Using the identity cosec² θ - cot² θ = 1, we get cot² θ = cosec² θ - 1, so cot² θ + cosec² θ + 1 + 2 cosec θ = (cosec² θ - 1) + cosec² θ + 1 + 2 cosec θ = 2 cosec² θ + 2 cosec θ = 2 cosec θ(cosec θ + 1). The denominator is cot θ(cosec θ + 1). Thus, the fraction becomes 2 cosec θ(cosec θ + 1) / [cot θ(cosec θ + 1)] = 2 cosec θ / cot θ = 2 × (1 / sin θ) × (sin θ / cos θ) = 2 / cos θ = 2 sec θ. Hence proved.
In simple words: Add the two fractions. Apply the identity cosec² θ - cot² θ = 1 to the numerator. Cancel the common (cosec θ + 1) factor. Simplify to get 2 sec θ.

Exam Tip: Use the identity cosec² θ - cot² θ = 1 to rewrite one variable in terms of the other - this reduces the complexity of the algebra.

Question 5(iii). Prove the following trigonometry identity: (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ
Answer: Working from the left side, we have \( (sin θ + cos θ)(cosec θ - sec θ) \). Substituting the reciprocal definitions, this becomes \( (sin θ + cos θ) × ( \frac{1}{sin θ} - \frac{1}{cos θ} ) \). Simplifying the bracket yields \( (sin θ + cos θ) × \frac{cos θ - sin θ}{sin θ cos θ} \). Expanding gives \( \frac{sin θ cos θ - sin^2 θ + cos^2 θ - sin θ cos θ}{sin θ cos θ} \), which reduces to \( \frac{cos^2 θ - sin^2 θ}{sin θ cos θ} \). Using the identity \( cos^2 θ = 1 - sin^2 θ \), this becomes \( \frac{1 - 2 sin^2 θ}{sin θ cos θ} \). Breaking this into two fractions, we get \( \frac{1}{sin θ cos θ} - \frac{2 sin^2 θ}{sin θ cos θ} \), which simplifies to \( cosec θ sec θ - 2 tan θ \), proving the identity.
In simple words: Start with the left side. Change cosec and sec to their reciprocal forms, then multiply out and simplify using basic trig identities until you reach the right side.

Exam Tip: Always convert cosec and sec to fractions with sin and cos right at the start. Group like terms carefully when expanding, and use \( sin^2 θ + cos^2 θ = 1 \) to simplify.

Question 6(i). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: sec⁴ A(1 - sin⁴ A) - 2 tan² A = 1
Answer: Starting with the left side, expand \( sec^4 A(1 - sin^4 A) \) as \( \frac{1}{cos^4 A} × (1 - sin^4 A) \). Factor \( 1 - sin^4 A = (1 + sin^2 A)(1 - sin^2 A) = (1 + sin^2 A)cos^2 A \). Substituting yields \( \frac{(1 + sin^2 A)cos^2 A}{cos^4 A} = \frac{1 + sin^2 A}{cos^2 A} \). Now subtract \( 2 tan^2 A = \frac{2 sin^2 A}{cos^2 A} \), giving \( \frac{1 + sin^2 A - 2 sin^2 A}{cos^2 A} = \frac{1 - sin^2 A}{cos^2 A} \). Since \( 1 - sin^2 A = cos^2 A \), we have \( \frac{cos^2 A}{cos^2 A} = 1 \), completing the proof.
In simple words: Use the difference of squares to factor \( 1 - sin^4 A \). Then carefully combine all the fractions and use the main identity \( sin^2 A + cos^2 A = 1 \) to finish.

Exam Tip: Recognize \( 1 - sin^4 A \) as a difference of squares immediately - this is the key factorization. Watch your arithmetic when combining the final fractions.

Question 6(ii). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: \( \frac{1}{sin A + cos A + 1} + \frac{1}{sin A + cos A - 1} = sec A + cosec A \)
Answer: Take the left side and find a common denominator, combining the two fractions into \( \frac{(sin A + cos A - 1) + (sin A + cos A + 1)}{(sin A + cos A + 1)(sin A + cos A - 1)} \). The numerator simplifies to \( 2(sin A + cos A) \). For the denominator, use the difference of squares pattern: \( (sin A + cos A + 1)(sin A + cos A - 1) = (sin A + cos A)^2 - 1 = sin^2 A + cos^2 A + 2 sin A cos A - 1 = 2 sin A cos A \). So the fraction becomes \( \frac{2(sin A + cos A)}{2 sin A cos A} = \frac{sin A}{sin A cos A} + \frac{cos A}{sin A cos A} = cosec A + sec A \), confirming the identity.
In simple words: Add the two fractions by finding a common denominator. Use the difference of squares formula in the denominator, then split the result into two simpler fractions.

Exam Tip: Recognize the denominator as a difference-of-squares pattern immediately. Expand \( (sin A + cos A)^2 \) carefully, and remember that \( sin^2 A + cos^2 A = 1 \).

Question 7(i). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: \( \frac{sin^3 θ + cos^3 θ}{sin θ + cos θ} + sin θ cos θ = 1 \)
Answer: Start with the left side. Factor the numerator using the sum of cubes formula: \( sin^3 θ + cos^3 θ = (sin θ + cos θ)(sin^2 θ + cos^2 θ - sin θ cos θ) \). Substituting into the fraction yields \( \frac{(sin θ + cos θ)(sin^2 θ + cos^2 θ - sin θ cos θ)}{sin θ + cos θ} + sin θ cos θ \). The \( (sin θ + cos θ) \) terms cancel, leaving \( (sin^2 θ + cos^2 θ - sin θ cos θ) + sin θ cos θ \). The \( sin θ cos θ \) terms cancel, and since \( sin^2 θ + cos^2 θ = 1 \), we get \( 1 \), confirming the identity.
In simple words: Use the sum of cubes formula to factor the numerator. Cancel the common terms in the fraction, then the \( sin θ cos θ \) parts cancel too, leaving just 1.

Exam Tip: Know the sum of cubes factorization \( a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \) - this is essential for this problem. After canceling, make sure all terms combine properly.

Question 7(ii). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: \( \frac{sin θ + cos θ}{sin θ + cos θ} + sin θ cos θ = 1 \)
Answer: (This question appears identical to Question 7(i) in structure. Using the same approach: factor \( sin^3 θ + cos^3 θ = (sin θ + cos θ)(sin^2 θ - sin θ cos θ + cos^2 θ) \). The first term simplifies to \( sin^2 θ - sin θ cos θ + cos^2 θ = 1 - sin θ cos θ \). Adding \( sin θ cos θ \) gives \( 1 - sin θ cos θ + sin θ cos θ = 1 \).
In simple words: Factor the sum of cubes, cancel, and simplify. The key identity \( sin^2 θ + cos^2 θ = 1 \) makes the final step straightforward.

Exam Tip: Be careful to recognize and correctly apply the sum of cubes formula. Always combine like terms (especially the \( sin θ cos θ \) terms) before concluding.

Question 8(i). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: (sec A - tan A)² (1 + sin A) = 1 - sin A
Answer: Start with the left side. Rewrite \( sec A - tan A = \frac{1}{cos A} - \frac{sin A}{cos A} = \frac{1 - sin A}{cos A} \). Squaring gives \( (\frac{1 - sin A}{cos A})^2 = \frac{(1 - sin A)^2}{cos^2 A} \). Multiplying by \( (1 + sin A) \) yields \( \frac{(1 - sin A)^2 (1 + sin A)}{cos^2 A} \). Using \( cos^2 A = 1 - sin^2 A = (1 - sin A)(1 + sin A) \), the denominator can be written as \( (1 - sin A)(1 + sin A) \). Therefore, \( \frac{(1 - sin A)^2 (1 + sin A)}{(1 - sin A)(1 + sin A)} = \frac{(1 - sin A)^2}{1} = 1 - sin A \), proving the identity.
In simple words: Combine sec and tan into a single fraction. Square it. Then multiply by \( (1 + sin A) \) and simplify using the Pythagorean identity.

Exam Tip: Express \( sec A - tan A \) as a single fraction right away. Recognize that \( cos^2 A = (1 - sin A)(1 + sin A) \) to make the cancellation clean.

Question 8(ii). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: \( \frac{cos A}{1 - tan A} - \frac{sin^2 A}{cos A - sin A} = sin A + cos A \)
Answer: For the first fraction, rewrite \( 1 - tan A = 1 - \frac{sin A}{cos A} = \frac{cos A - sin A}{cos A} \), so \( \frac{cos A}{1 - tan A} = \frac{cos A}{\frac{cos A - sin A}{cos A}} = \frac{cos^2 A}{cos A - sin A} \). Now combine both fractions: \( \frac{cos^2 A}{cos A - sin A} - \frac{sin^2 A}{cos A - sin A} = \frac{cos^2 A - sin^2 A}{cos A - sin A} \). Factor the numerator as \( (cos A - sin A)(cos A + sin A) \), giving \( \frac{(cos A - sin A)(cos A + sin A)}{cos A - sin A} = cos A + sin A \), proving the identity.
In simple words: Rewrite the denominator \( 1 - tan A \) as a fraction with sin and cos. Get a common denominator. Then factor the numerator and cancel.

Exam Tip: Convert \( tan A \) to \( \frac{sin A}{cos A} \) immediately. Combine the fractions carefully, and factor the difference of squares in the numerator.

Question 8(iii). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: (sec A - cosec A)(1 + tan A + cot A) = tan A sec A - cot A cosec A
Answer: Rewrite the left side: \( sec A - cosec A = \frac{1}{cos A} - \frac{1}{sin A} = \frac{sin A - cos A}{sin A cos A} \) and \( 1 + tan A + cot A = 1 + \frac{sin A}{cos A} + \frac{cos A}{sin A} = \frac{sin A cos A + sin^2 A + cos^2 A}{sin A cos A} = \frac{sin A cos A + 1}{sin A cos A} \). Multiplying these gives \( \frac{(sin A - cos A)(sin A cos A + 1)}{sin^2 A cos^2 A} \). For the right side: \( tan A sec A = \frac{sin A}{cos A} × \frac{1}{cos A} = \frac{sin A}{cos^2 A} \) and \( cot A cosec A = \frac{cos A}{sin A} × \frac{1}{sin A} = \frac{cos A}{sin^2 A} \). So \( tan A sec A - cot A cosec A = \frac{sin^3 A - cos^3 A}{sin^2 A cos^2 A} = \frac{(sin A - cos A)(sin^2 A + sin A cos A + cos^2 A)}{sin^2 A cos^2 A} = \frac{(sin A - cos A)(1 + sin A cos A)}{sin^2 A cos^2 A} \), which matches the left side, completing the proof.
In simple words: Convert everything to sin and cos fractions. Multiply out the left side and simplify the right side separately, then show they are equal.

Exam Tip: Work each side independently. Recognize the sum of cubes factor \( sin^3 A - cos^3 A = (sin A - cos A)(sin^2 A + sin A cos A + cos^2 A) \) on the right side.

Question 8(iv). Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: \( \frac{tan^2 θ}{tan^2 θ - 1} + \frac{cosec^2 θ}{sec^2 θ - cosec^2 θ} = \frac{1}{sin^2 θ - cos^2 θ} \)
Answer: For the first fraction, substitute \( tan^2 θ = \frac{sin^2 θ}{cos^2 θ} \) and simplify the denominator: \( tan^2 θ - 1 = \frac{sin^2 θ - cos^2 θ}{cos^2 θ} \), so the first fraction becomes \( \frac{\frac{sin^2 θ}{cos^2 θ}}{\frac{sin^2 θ - cos^2 θ}{cos^2 θ}} = \frac{sin^2 θ}{sin^2 θ - cos^2 θ} \). For the second fraction, \( sec^2 θ - cosec^2 θ = \frac{1}{cos^2 θ} - \frac{1}{sin^2 θ} = \frac{sin^2 θ - cos^2 θ}{sin^2 θ cos^2 θ} \), so the second fraction becomes \( \frac{cosec^2 θ}{\frac{sin^2 θ - cos^2 θ}{sin^2 θ cos^2 θ}} = \frac{\frac{1}{sin^2 θ} × sin^2 θ cos^2 θ}{sin^2 θ - cos^2 θ} = \frac{cos^2 θ}{sin^2 θ - cos^2 θ} \). Adding both fractions gives \( \frac{sin^2 θ + cos^2 θ}{sin^2 θ - cos^2 θ} = \frac{1}{sin^2 θ - cos^2 θ} \), confirming the identity.
In simple words: Convert tan, sec, and cosec to sin and cos forms. Simplify each fraction separately, then add them using a common denominator.

Exam Tip: Work with one fraction at a time. Convert all reciprocal ratios to sin and cos immediately. The key is recognizing that both fractions share a common denominator after simplification.

Question 9. If \( \frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{sin^2 A - cos^2 A} \), then prove that \( \frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{1 - 2cos^2 A} = \frac{2 sec^2 A}{tan^2 A - 1} \)
Answer: Start by finding a common denominator for the left side: \( \frac{(sin A + cos A)^2 + (sin A - cos A)^2}{(sin A - cos A)(sin A + cos A)} \). Expand the numerator: \( (sin A + cos A)^2 + (sin A - cos A)^2 = sin^2 A + 2 sin A cos A + cos^2 A + sin^2 A - 2 sin A cos A + cos^2 A = 2(sin^2 A + cos^2 A) = 2 \). For the denominator, use the difference of squares: \( (sin A - cos A)(sin A + cos A) = sin^2 A - cos^2 A \). So the left side equals \( \frac{2}{sin^2 A - cos^2 A} \). Now simplify this to the other forms. Since \( sin^2 A = 1 - cos^2 A \), we have \( sin^2 A - cos^2 A = 1 - cos^2 A - cos^2 A = 1 - 2 cos^2 A \), so \( \frac{2}{sin^2 A - cos^2 A} = \frac{2}{1 - 2 cos^2 A} \). Dividing numerator and denominator by \( cos^2 A \) yields \( \frac{\frac{2}{cos^2 A}}{\frac{1 - 2 cos^2 A}{cos^2 A}} = \frac{2 sec^2 A}{tan^2 A - 1} \), completing the proof.
In simple words: Add the two fractions by finding a common denominator. Expand the squares in the numerator and simplify. Then convert between the different forms using basic trig identities.

Exam Tip: Expand \( (sin A ± cos A)^2 \) carefully - the cross terms cancel when added. Always recognize \( a^2 - b^2 = (a-b)(a+b) \) for quick factorization of the denominator.

Question 10. Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined: 2(sin⁶ θ + cos⁶ θ) - 3(sin⁴ θ + cos⁴ θ) + 1 = 0
Answer: Rewrite the expression as \( 2[(sin^2 θ)^3 + (cos^2 θ)^3] - 3[(sin^2 θ)^2 + (cos^2 θ)^2] + 1 \). Use the sum of cubes formula: \( sin^6 θ + cos^6 θ = (sin^2 θ + cos^2 θ)^3 - 3 sin^2 θ cos^2 θ(sin^2 θ + cos^2 θ) = 1 - 3 sin^2 θ cos^2 θ \). For the even fourth powers: \( sin^4 θ + cos^4 θ = (sin^2 θ + cos^2 θ)^2 - 2 sin^2 θ cos^2 θ = 1 - 2 sin^2 θ cos^2 θ \). Substituting both yields \( 2(1 - 3 sin^2 θ cos^2 θ) - 3(1 - 2 sin^2 θ cos^2 θ) + 1 = 2 - 6 sin^2 θ cos^2 θ - 3 + 6 sin^2 θ cos^2 θ + 1 = 2 - 3 + 1 = 0 \), confirming the identity.
In simple words: Express the sixth powers and fourth powers in terms of \( sin^2 θ \) and \( cos^2 θ \). Use the sum formulas to simplify, then substitute and collect like terms to show the result is zero.

Exam Tip: Recognize that you can write \( sin^6 θ + cos^6 θ \) using the sum of cubes formula for \( (sin^2 θ)^3 + (cos^2 θ)^3 \). The \( sin^2 θ cos^2 θ \) terms will cancel out perfectly at the end.

Question 11. If cot θ + cos θ = m and cot θ - cos θ = n, then prove that (m² - n²)² = 16 mn
Answer: Given the two equations, add them: \( m + n = 2 cot θ \), so \( cot θ = \frac{m + n}{2} \) and \( tan θ = \frac{2}{m + n} \). Subtract the second from the first: \( m - n = 2 cos θ \), so \( cos θ = \frac{m - n}{2} \) and \( sec θ = \frac{2}{m - n} \). Square both results and use the identity \( sec^2 θ - tan^2 θ = 1 \): \( (\frac{2}{m - n})^2 - (\frac{2}{m + n})^2 = 1 \). This simplifies to \( \frac{4}{(m - n)^2} - \frac{4}{(m + n)^2} = 1 \). Multiplying by the common denominator yields \( 4[(m + n)^2 - (m - n)^2] = (m - n)^2 (m + n)^2 \). Expand the left side: \( (m + n)^2 - (m - n)^2 = [(m + n) - (m - n)][(m + n) + (m - n)] = (2n)(2m) = 4mn \), so \( 4 × 4mn = (m^2 - n^2)^2 \), giving \( 16mn = (m^2 - n^2)^2 \), completing the proof.
In simple words: Add and subtract the two given equations to find cot and cos in terms of m and n. Convert to sec and tan, then use the Pythagorean identity to create an equation. Solve algebraically to show the desired relationship.

Exam Tip: The key step is adding and subtracting the given equations. Use the Pythagorean identity \( sec^2 θ - tan^2 θ = 1 \) confidently. When simplifying, recognize the difference-of-squares pattern.

Question 12(i). When 0° < θ < 90°, solve the following equation: 2 cos² θ + sin θ - 2 = 0
Answer: Substitute \( cos^2 θ = 1 - sin^2 θ \) to get \( 2(1 - sin^2 θ) + sin θ - 2 = 0 \), which simplifies to \( 2 - 2 sin^2 θ + sin θ - 2 = 0 \) or \( sin θ - 2 sin^2 θ = 0 \). Factor to obtain \( sin θ (1 - 2 sin θ) = 0 \). This gives either \( sin θ = 0 \) or \( 1 - 2 sin θ = 0 \). If \( sin θ = 0 \), then \( θ = 0° \), but since \( θ > 0° \), this is rejected. If \( 1 - 2 sin θ = 0 \), then \( sin θ = \frac{1}{2} = sin 30° \), so \( θ = 30° \). Therefore, the answer is \( θ = 30° \).
In simple words: Use the Pythagorean identity to rewrite in terms of sin alone. Factor the resulting equation. Check which solution satisfies the given restriction on θ.

Exam Tip: Always convert everything to a single trig function using identities. Check your solutions against the given domain restriction - \( θ > 0° \) rules out \( θ = 0° \).

Question 12(ii). When 0° < θ < 90°, solve the following equation: 3 cos θ = 2 sin² θ
Answer: Substitute \( sin^2 θ = 1 - cos^2 θ \) to get \( 3 cos θ = 2(1 - cos^2 θ) \), which expands to \( 3 cos θ = 2 - 2 cos^2 θ \) or \( 2 cos^2 θ + 3 cos θ - 2 = 0 \). Factor by splitting the middle term: \( 2 cos^2 θ + 4 cos θ - cos θ - 2 = 0 \) becomes \( 2 cos θ(cos θ + 2) - 1(cos θ + 2) = 0 \) or \( (2 cos θ - 1)(cos θ + 2) = 0 \). This gives \( cos θ = \frac{1}{2} \) or \( cos θ = -2 \). Since \( cos θ \) must be between -1 and 1, the solution \( cos θ = -2 \) is impossible. From \( cos θ = \frac{1}{2} = cos 60° \), we get \( θ = 60° \). Therefore, the answer is \( θ = 60° \).
In simple words: Replace sin² with \( 1 - cos^2 θ \). Rearrange to get a quadratic in cos θ. Factor and solve, discarding impossible values of cos θ.

Exam Tip: Remember that cosine values must lie between -1 and 1 inclusive. When you get \( cos θ = -2 \), reject it immediately. Quadratic factorization is essential here.

Question 12(iii). When 0° < θ < 90°, solve the following equation: sec² θ - 2 tan θ = 0
Answer: Use the identity \( sec^2 θ = 1 + tan^2 θ \) to rewrite as \( 1 + tan^2 θ - 2 tan θ = 0 \) or \( tan^2 θ - 2 tan θ + 1 = 0 \). This is a perfect square: \( (tan θ - 1)^2 = 0 \), giving \( tan θ = 1 \). Since \( tan 45° = 1 \), we have \( θ = 45° \). Therefore, the answer is \( θ = 45° \).
In simple words: Replace \( sec^2 θ \) with \( 1 + tan^2 θ \). Rearrange into a quadratic. Recognize it as a perfect square and solve.

Exam Tip: Recognize perfect square trinomials immediately - they save time. Know that \( tan 45° = 1 \) is a standard angle value.

Question 12(iv). When 0° < θ < 90°, solve the following equation: (Equation not displayed in source)
Answer: (Question text was not visible in the provided excerpt. The corresponding answer working should follow the same pattern: convert to a single trig function using identities, solve the resulting equation, and check against the domain constraint.)

Exam Tip: Always use the Pythagorean identities to convert to a single trig function before solving. Verify that your angle falls within the given range.

Question. Solve for θ: tan² θ = 3 (sec θ - 1).
Answer: We start with the given equation tan² θ = 3 (sec θ - 1). Using the identity tan² θ = sec² θ - 1, we substitute to get sec² θ - 1 = 3 sec θ - 3. Rearranging this becomes sec² θ - 3 sec θ + 2 = 0. Factoring the left side, we write this as sec² θ - 2 sec θ - sec θ + 2 = 0, which factors to (sec θ - 1)(sec θ - 2) = 0. This gives us sec θ = 1 or sec θ = 2. When sec θ = 1, we have θ = 0°, but since we are told θ > 0°, this solution is rejected. When sec θ = 2, we have sec θ = sec 60°, so θ = 60°. Therefore, θ = 60°.
In simple words: Use the tan-sec identity to turn the equation into a quadratic in sec θ. Factor it to find two possible values, then pick the one that satisfies the given constraint θ > 0°.

Exam Tip: Always check boundary conditions - here, θ = 0° must be rejected because the problem states θ > 0°. Examiners reward answers that show this reasoning explicitly.