GSEB Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.4

Question 1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. (CBSE 2012)
Answer: We need to express sin A, sec A, and tan A using cot A.
First, for sin A:
\( \sin A = \frac{ 1 }{ \text{cosec A} } \)
We know the identity \( \text{cosec}^2 A = 1 + \text{cot}^2 A \), so \( \text{cosec A} = \sqrt{1 + \text{cot}^2 A} \).
Therefore, \( \sin A = \frac{ 1 }{ \sqrt{1 + \text{cot}^2 A} } \)

Next, for sec A:
We know the identity \( \text{sec}^2 A = 1 + \text{tan}^2 A \), and \( \text{tan A} = \frac{ 1 }{ \text{cot A} } \).
So, \( \text{sec A} = \sqrt{1 + \text{tan}^2 A} = \sqrt{1 + \frac{ 1 }{ \text{cot}^2 A }} \)
\( \text{sec A} = \sqrt{\frac{ \text{cot}^2 A + 1 }{ \text{cot}^2 A }} = \frac{ \sqrt{1 + \text{cot}^2 A} }{ \text{cot A} } \)

Lastly, for tan A:
The relationship is direct: \( \text{tan A} = \frac{ 1 }{ \text{cot A} } \)
In simple words: To write sin A, sec A, and tan A using cot A, we use known mathematical rules and formulas. For sin A, we turn it into cosec A, then to cot A. For sec A, we use tan A and then cot A. For tan A, it's a simple flip with cot A.

Exam Tip: Always recall the fundamental trigonometric identities (like \( \text{sin}^2 A + \text{cos}^2 A = 1 \), \( 1 + \text{tan}^2 A = \text{sec}^2 A \), and \( 1 + \text{cot}^2 A = \text{cosec}^2 A \)) when trying to express one ratio in terms of another.

Question 2. Write all the other trigonometric ratios of \( \angle A \) in terms of sec A.
Answer: We need to express sin A, cos A, tan A, cosec A, and cot A using sec A.
First, for cos A:
The reciprocal relationship is simple: \( \text{cos A} = \frac{ 1 }{ \text{sec A} } \)

Next, for sin A:
We know the identity \( \sin^2 A + \cos^2 A = 1 \).
So, \( \sin^2 A = 1 - \cos^2 A \)
Substituting \( \cos A = \frac{ 1 }{ \text{sec A} } \):
\( \sin^2 A = 1 - \frac{ 1 }{ \text{sec}^2 A } = \frac{ \text{sec}^2 A - 1 }{ \text{sec}^2 A } \)
Therefore, \( \sin A = \sqrt{\frac{ \text{sec}^2 A - 1 }{ \text{sec}^2 A }} = \frac{ \sqrt{\text{sec}^2 A - 1} }{ \text{sec A} } \)

For tan A:
We know the identity \( \text{sec}^2 A = 1 + \text{tan}^2 A \).
So, \( \text{tan}^2 A = \text{sec}^2 A - 1 \)
Therefore, \( \text{tan A} = \sqrt{\text{sec}^2 A - 1} \)

For cosec A:
This is the reciprocal of sin A:
\( \text{cosec A} = \frac{ 1 }{ \sin A } = \frac{ 1 }{ \frac{ \sqrt{\text{sec}^2 A - 1} }{ \text{sec A} } } = \frac{ \text{sec A} }{ \sqrt{\text{sec}^2 A - 1} } \)

Lastly, for cot A:
This is the reciprocal of tan A:
\( \text{cot A} = \frac{ 1 }{ \text{tan A} } = \frac{ 1 }{ \sqrt{\text{sec}^2 A - 1} } \)
In simple words: To write all the other ratios using sec A, we use their basic definitions and mathematical rules. Cos A is simple, just 1 over sec A. For sin A and tan A, we use the square identities and then take the square root. Cosec A and cot A are then found by flipping the answers for sin A and tan A respectively.

Exam Tip: Remember to express all intermediate ratios (like sin A or tan A) in terms of the target ratio (sec A) before finding their reciprocals. This helps in maintaining consistency.

Question 3.
(i) \( \frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}} \)
(ii) \( \sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ} \)
Answer:
(i) To solve this, we use the complementary angle identities \( \sin (90^{\circ} - \theta) = \cos \theta \) and \( \cos (90^{\circ} - \theta) = \sin \theta \).
\( \frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}} \)
We can rewrite \( \sin 63^{\circ} \) as \( \sin (90^{\circ} - 27^{\circ}) = \cos 27^{\circ} \).
Also, rewrite \( \cos 73^{\circ} \) as \( \cos (90^{\circ} - 17^{\circ}) = \sin 17^{\circ} \).
Substituting these into the expression:
\( = \frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\sin ^{2} 17^{\circ}} \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \frac{1}{1} = 1 \)

(ii) For this part, we again use complementary angle identities.
\( \sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ} \)
Rewrite \( \cos 65^{\circ} \) as \( \cos (90^{\circ} - 25^{\circ}) = \sin 25^{\circ} \).
Rewrite \( \sin 65^{\circ} \) as \( \sin (90^{\circ} - 25^{\circ}) = \cos 25^{\circ} \).
Substituting these values:
\( = \sin 25^{\circ} \sin 25^{\circ} + \cos 25^{\circ} \cos 25^{\circ} \)
\( = \sin^2 25^{\circ} + \cos^2 25^{\circ} \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = 1 \)
In simple words: For both parts, we change one of the angles so that it matches the other, by using the rule that sin of an angle is cos of its complement (and vice versa). Then, we apply the basic rule \( \sin^2 \theta + \cos^2 \theta = 1 \), which makes both expressions simplify to 1.

Exam Tip: Look for angles that add up to 90 degrees. This is a strong indicator that you should use complementary angle identities like \( \sin(90^{\circ}-\theta) = \cos \theta \) to simplify the expression and often lead to \( \sin^2 \theta + \cos^2 \theta = 1 \).

Question 4. Choose the correct option. Justify your choice.
(i) \( 9 \sec^2 \theta - 9 \tan^2 \theta = \)
(a) 9
(b) 1
(c) 8
(d) 0
Answer: (a) 9

Solution:
Given expression: \( 9 \sec^2 \theta - 9 \tan^2 \theta \)
We can take out 9 as a common factor:
\( = 9 (\sec^2 \theta - \tan^2 \theta) \)
We know the trigonometric identity \( \sec^2 \theta - \tan^2 \theta = 1 \).
Substituting this into the expression:
\( = 9 \times 1 \)
\( = 9 \)
So, the correct option is (a).
In simple words: When you see \( \sec^2 \theta - \tan^2 \theta \), remember it always equals 1. So, if there's a 9 in front, the answer is simply 9 times 1, which is 9.

Exam Tip: Always factor out common terms first. Recognizing the identity \( \sec^2 \theta - \tan^2 \theta = 1 \) is crucial for solving such problems quickly.

Question 4. Choose the correct option. Justify your choice.
(ii) \( (1 + \tan \theta + \sec \theta) (1 + \cot \theta - \text{cosec } \theta) = \)
(a) -1
(b) 1
(c) 2
(d) -1
Answer: (c) 2

Solution:
Given expression: \( (1 + \tan \theta + \sec \theta) (1 + \cot \theta - \text{cosec } \theta) \)
Convert all terms into \( \sin \theta \) and \( \cos \theta \):
\( = \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) \)
Combine terms inside each parenthesis:
\( = \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \)
Let \( A = \cos \theta + \sin \theta \) and \( B = 1 \). The expression now looks like \( \frac{(A + B)(A - B)}{\cos \theta \sin \theta} \).
Using the identity \( (X+Y)(X-Y) = X^2 - Y^2 \):
\( = \frac{(\cos \theta + \sin \theta)^2 - 1^2}{\cos \theta \sin \theta} \)
Expand \( (\cos \theta + \sin \theta)^2 \):
\( = \frac{\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta - 1}{\cos \theta \sin \theta} \)
We know \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( = \frac{1 + 2 \sin \theta \cos \theta - 1}{\cos \theta \sin \theta} \)
\( = \frac{2 \sin \theta \cos \theta}{\cos \theta \sin \theta} \)
Cancel out \( \sin \theta \cos \theta \):
\( = 2 \)
Thus, the correct option is (c).
In simple words: Convert everything to sin and cos. Group the terms \( (\sin \theta + \cos \theta) \) and \( 1 \) to use the \( (A+B)(A-B) \) rule. Then, simplify the numerator using \( \sin^2 \theta + \cos^2 \theta = 1 \) to get a simple result of 2.

Exam Tip: When faced with expressions involving different trigonometric ratios, converting them all to sine and cosine terms is a common and effective strategy. Also, look for patterns like \( (A+B)(A-B) \) to simplify complex multiplications.

Question 4. Choose the correct option. Justify your choice.
(iii) \( (\sec A + \tan A) (1 - \sin A) = \)
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Answer: (d) cos A

Solution:
Given expression: \( (\sec A + \tan A) (1 - \sin A) \)
Convert \( \sec A \) and \( \tan A \) into \( \sin A \) and \( \cos A \):
\( = \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 - \sin A) \)
Combine terms inside the first parenthesis:
\( = \left(\frac{1 + \sin A}{\cos A}\right) (1 - \sin A) \)
Multiply the numerators:
\( = \frac{(1 + \sin A)(1 - \sin A)}{\cos A} \)
Using the identity \( (X+Y)(X-Y) = X^2 - Y^2 \):
\( = \frac{1^2 - \sin^2 A}{\cos A} \)
We know that \( 1 - \sin^2 A = \cos^2 A \):
\( = \frac{\cos^2 A}{\cos A} \)
Cancel out one \( \cos A \):
\( = \cos A \)
So, the correct option is (d).
In simple words: Change sec A and tan A to their sin and cos forms. Combine them, then multiply by \( (1 - \sin A) \). Use the \( (A+B)(A-B) \) rule and the \( \sin^2 A + \cos^2 A = 1 \) rule to simplify, leaving only cos A.

Exam Tip: Similar to the previous question, converting all ratios to sine and cosine is a good starting point. Look for opportunities to use algebraic identities like \( (a+b)(a-b) = a^2-b^2 \) along with trigonometric identities.

Question 4. Choose the correct option. Justify your choice.
(iv) \( \frac{1+\tan ^{2} A}{1+\cot ^{2} A} = \)
(a) sec 2A
(b) -1
(c) cot 2A
(d) tan² A
Answer: (d) tan² A

Solution:
Given expression: \( \frac{1+\tan ^{2} A}{1+\cot ^{2} A} \)
We know the trigonometric identities:
\( 1 + \tan^2 A = \sec^2 A \)
\( 1 + \cot^2 A = \text{cosec}^2 A \)
Substitute these identities into the expression:
\( = \frac{\sec^2 A}{\text{cosec}^2 A} \)
Now, convert \( \sec^2 A \) and \( \text{cosec}^2 A \) into \( \sin A \) and \( \cos A \):
\( \sec^2 A = \frac{1}{\cos^2 A} \)
\( \text{cosec}^2 A = \frac{1}{\sin^2 A} \)
So the expression becomes:
\( = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \)
\( = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} \)
\( = \frac{\sin^2 A}{\cos^2 A} \)
We know that \( \frac{\sin A}{\cos A} = \tan A \), so \( \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \).
Thus, \( = \tan^2 A \)
So, the correct option is (d).
In simple words: Replace \( (1 + \tan^2 A) \) with \( \sec^2 A \) and \( (1 + \cot^2 A) \) with \( \text{cosec}^2 A \). Then, change sec and cosec into their cos and sin forms. Simplify the fraction, and you'll get \( \tan^2 A \).

Exam Tip: Master the reciprocal and Pythagorean identities, as they are fundamental for simplifying trigonometric expressions. Remember that \( \frac{\sec^2 A}{\text{cosec}^2 A} \) directly simplifies to \( \tan^2 A \).

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(i) \( (\text{cosec } \theta - \cot \theta)^2 = \frac{1-\cos \theta}{1+\cos \theta} \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = (\text{cosec } \theta - \cot \theta)^2 \)
Convert \( \text{cosec } \theta \) and \( \cot \theta \) into \( \sin \theta \) and \( \cos \theta \):
\( = \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 \)
Combine the terms inside the parenthesis:
\( = \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 \)
Apply the square to both numerator and denominator:
\( = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \)
We know the identity \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this into the denominator:
\( = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} \)
The denominator \( 1 - \cos^2 \theta \) can be factored as \( (1 - \cos \theta)(1 + \cos \theta) \) using \( A^2 - B^2 = (A-B)(A+B) \).
\( = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \)
Cancel out one \( (1 - \cos \theta) \) term from the numerator and denominator:
\( = \frac{1 - \cos \theta}{1 + \cos \theta} \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: Begin by changing cosec and cot into sin and cos. Simplify the fraction inside the bracket and then square it. Use the identity for \( \sin^2 \theta \) in the denominator and factor it, then cancel out common terms to reach the right side of the equation.

Exam Tip: When proving identities, always try to simplify the more complex side (usually LHS) first. Converting all trigonometric ratios to sine and cosine is a reliable first step. Also, remember to look for algebraic identities like \( A^2-B^2 = (A-B)(A+B) \).

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(ii) \( \frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A} = 2 \sec A \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = \frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A} \)
Combine the two fractions by finding a common denominator, which is \( (1+\sin A)(\cos A) \):
\( = \frac{\cos A \cdot \cos A + (1+\sin A) \cdot (1+\sin A)}{(1+\sin A)(\cos A)} \)
\( = \frac{\cos^2 A + (1+\sin A)^2}{\cos A (1+\sin A)} \)
Expand \( (1+\sin A)^2 \), which is \( 1^2 + 2(1)(\sin A) + \sin^2 A = 1 + 2\sin A + \sin^2 A \):
\( = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{\cos A (1+\sin A)} \)
Rearrange the terms in the numerator:
\( = \frac{(\cos^2 A + \sin^2 A) + 1 + 2\sin A}{\cos A (1+\sin A)} \)
We know the identity \( \cos^2 A + \sin^2 A = 1 \). Substitute this into the numerator:
\( = \frac{1 + 1 + 2\sin A}{\cos A (1+\sin A)} \)
\( = \frac{2 + 2\sin A}{\cos A (1+\sin A)} \)
Factor out 2 from the numerator:
\( = \frac{2 (1 + \sin A)}{\cos A (1+\sin A)} \)
Cancel out the common term \( (1+\sin A) \) from the numerator and denominator:
\( = \frac{2}{\cos A} \)
We know that \( \frac{1}{\cos A} = \sec A \).
\( = 2 \sec A \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: Start by adding the two fractions together using a common denominator. Expand the squared term and use the identity \( \sin^2 A + \cos^2 A = 1 \). Simplify the numerator, factor out 2, and then cancel common parts. What remains is 2 times \( \frac{1}{\cos A} \), which is \( 2 \sec A \).

Exam Tip: For problems involving sums or differences of fractions, always aim to combine them into a single fraction first by finding a common denominator. This usually reveals opportunities to apply identities like \( \sin^2 A + \cos^2 A = 1 \) or simplify terms.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(iii) \( \frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} = 1 + \sec \theta \text{cosec } \theta \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = \frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} \)
Convert \( \tan \theta \) and \( \cot \theta \) into \( \sin \theta \) and \( \cos \theta \):
\( = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} \)
Simplify the denominators:
\( = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}} \)
Now, invert and multiply:
\( = \frac{\sin \theta}{\cos \theta} \cdot \frac{\sin \theta}{\sin \theta - \cos \theta} + \frac{\cos \theta}{\sin \theta} \cdot \frac{\cos \theta}{\cos \theta - \sin \theta} \)
\( = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)} \)
Notice that \( (\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta) \). Let's make the denominators uniform:
\( = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{- \sin \theta (\sin \theta - \cos \theta)} \)
\( = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)} \)
Now combine the fractions with a common denominator \( \sin \theta \cos \theta (\sin \theta - \cos \theta) \):
\( = \frac{\sin \theta \cdot \sin^2 \theta - \cos \theta \cdot \cos^2 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
\( = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
Use the algebraic identity \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \):
\( = \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
Cancel out \( (\sin \theta - \cos \theta) \) from numerator and denominator:
\( = \frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
We know \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \)
Separate the fraction:
\( = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( = \frac{1}{\sin \theta \cos \theta} + 1 \)
We know \( \frac{1}{\sin \theta} = \text{cosec } \theta \) and \( \frac{1}{\cos \theta} = \sec \theta \).
\( = \text{cosec } \theta \sec \theta + 1 \)
\( = 1 + \sec \theta \text{cosec } \theta \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: First, rewrite tangent and cotangent in terms of sine and cosine. Simplify the complex fractions. Adjust the denominators so they are the same by factoring out a negative sign. Combine the fractions, then use the \( a^3 - b^3 \) algebraic rule to simplify the top part. After canceling terms and using \( \sin^2 \theta + \cos^2 \theta = 1 \), you can split the fraction and convert it back to secant and cosecant.

Exam Tip: For complex fractions, convert all terms to sine and cosine. Always look for ways to make denominators common and pay attention to signs when factoring (e.g., \( (b-a) = -(a-b) \)). The \( a^3 \pm b^3 \) identities are sometimes useful in these types of proofs.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(iv) \( \frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A} \)
Answer:
Let's simplify both the Left Hand Side (LHS) and Right Hand Side (RHS) separately.

**Left Hand Side (LHS):**
\( \text{LHS} = \frac{1+\sec A}{\sec A} \)
Substitute \( \sec A = \frac{1}{\cos A} \):
\( = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} \)
Simplify the numerator:
\( = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} \)
Multiply the numerator by the reciprocal of the denominator:
\( = \frac{\cos A + 1}{\cos A} \times \frac{\cos A}{1} \)
Cancel out \( \cos A \):
\( = 1 + \cos A \)

**Right Hand Side (RHS):**
\( \text{RHS} = \frac{\sin^2 A}{1-\cos A} \)
We know the identity \( \sin^2 A = 1 - \cos^2 A \). Substitute this into the numerator:
\( = \frac{1 - \cos^2 A}{1-\cos A} \)
Factor the numerator using \( A^2 - B^2 = (A-B)(A+B) \), so \( 1 - \cos^2 A = (1-\cos A)(1+\cos A) \):
\( = \frac{(1-\cos A)(1+\cos A)}{1-\cos A} \)
Cancel out the common term \( (1-\cos A) \):
\( = 1 + \cos A \)
Since \( \text{LHS} = 1 + \cos A \) and \( \text{RHS} = 1 + \cos A \), then \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: For this proof, simplify both sides on their own. For the left side, change sec A to 1/cos A and then simplify the fraction. For the right side, change \( \sin^2 A \) to \( (1 - \cos^2 A) \), then factor the top part using the difference of squares rule and cancel out terms. Both sides should end up being the same.

Exam Tip: When both sides of an identity appear complex, simplifying both LHS and RHS independently until they match a common expression is an efficient approach. Always remember to convert to sine and cosine if unsure about other identities, and keep algebraic factoring in mind.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(v) \( \frac{\cos A-\sin A+1}{\cos A+\sin A-1} = \text{cosec } A + \cot A \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = \frac{\cos A-\sin A+1}{\cos A+\sin A-1} \)
To get terms of \( \text{cosec } A \) and \( \cot A \), divide the numerator and denominator by \( \sin A \):
\( = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} \)
Substitute \( \frac{\cos A}{\sin A} = \cot A \), \( \frac{\sin A}{\sin A} = 1 \), and \( \frac{1}{\sin A} = \text{cosec } A \):
\( = \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A} \)
Rearrange the numerator to group \( \cot A + \text{cosec } A \):
\( = \frac{(\cot A + \text{cosec } A) - 1}{\cot A - \text{cosec } A + 1} \)
Now, replace the \( 1 \) in the numerator using the identity \( \text{cosec}^2 A - \cot^2 A = 1 \):
\( = \frac{(\cot A + \text{cosec } A) - (\text{cosec}^2 A - \cot^2 A)}{\cot A - \text{cosec } A + 1} \)
Factor \( (\text{cosec}^2 A - \cot^2 A) \) as \( (\text{cosec } A - \cot A)(\text{cosec } A + \cot A) \):
\( = \frac{(\cot A + \text{cosec } A) - [(\text{cosec } A - \cot A)(\text{cosec } A + \cot A)]}{\cot A - \text{cosec } A + 1} \)
Factor out \( (\cot A + \text{cosec } A) \) from the numerator:
\( = \frac{(\cot A + \text{cosec } A) [1 - (\text{cosec } A - \cot A)]}{\cot A - \text{cosec } A + 1} \)
Simplify the term in the square brackets in the numerator:
\( = \frac{(\cot A + \text{cosec } A) [1 - \text{cosec } A + \cot A]}{\cot A - \text{cosec } A + 1} \)
Notice that the term in the square brackets \( (1 - \text{cosec } A + \cot A) \) is identical to the denominator \( (\cot A - \text{cosec } A + 1) \).
Cancel out the common term:
\( = \cot A + \text{cosec } A \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: To start, divide the entire top and bottom of the fraction by \( \sin A \) to get cot A and cosec A terms. Then, change the '1' in the numerator to \( (\text{cosec}^2 A - \cot^2 A) \) using a common identity. Factor this difference of squares. Finally, pull out the common term \( (\cot A + \text{cosec } A) \) from the numerator and cancel identical parts from the top and bottom to reach the desired answer.

Exam Tip: When the RHS involves cosecant and cotangent, dividing the LHS by \( \sin A \) (for all terms) is a helpful trick to introduce these ratios. The substitution of \( 1 = \text{cosec}^2 A - \cot^2 A \) is a common step in such proofs.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(vi) \( \sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = \sqrt{\frac{1+\sin A}{1-\sin A}} \)
To remove the square root from the denominator, multiply both the numerator and denominator inside the square root by \( (1+\sin A) \):
\( = \sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}} \)
\( = \sqrt{\frac{(1+\sin A)^2}{1^2 - \sin^2 A}} \)
In the denominator, use the identity \( 1 - \sin^2 A = \cos^2 A \):
\( = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} \)
Now, take the square root of both the numerator and the denominator:
\( = \frac{\sqrt{(1+\sin A)^2}}{\sqrt{\cos^2 A}} \)
\( = \frac{1+\sin A}{\cos A} \)
Split the fraction into two terms:
\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \)
We know that \( \frac{1}{\cos A} = \sec A \) and \( \frac{\sin A}{\cos A} = \tan A \).
\( = \sec A + \tan A \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: To remove the square root on the left side, multiply the top and bottom inside the root by \( (1+\sin A) \). This makes the top a perfect square and the bottom turns into \( \cos^2 A \). Take the square root, then separate the fraction into two parts, which will directly give you sec A and tan A.

Exam Tip: When you see a square root involving \( (1+\sin A)/(1-\sin A) \) or similar expressions, rationalize the denominator by multiplying by the conjugate. This is a standard technique to simplify such terms and reveal underlying identities.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(vii) \( \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta} = \tan \theta \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta} \)
Factor out \( \sin \theta \) from the numerator and \( \cos \theta \) from the denominator:
\( = \frac{\sin \theta (1-2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta-1)} \)
We know that \( \frac{\sin \theta}{\cos \theta} = \tan \theta \). So, the expression becomes:
\( = \tan \theta \cdot \frac{(1-2 \sin^2 \theta)}{(2 \cos^2 \theta-1)} \)
Now, we need to show that \( \frac{(1-2 \sin^2 \theta)}{(2 \cos^2 \theta-1)} = 1 \).
Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), which implies \( \sin^2 \theta = 1 - \cos^2 \theta \) and \( \cos^2 \theta = 1 - \sin^2 \theta \).
Substitute \( \sin^2 \theta = 1 - \cos^2 \theta \) into the numerator's bracket:
\( 1 - 2 \sin^2 \theta = 1 - 2(1 - \cos^2 \theta) = 1 - 2 + 2 \cos^2 \theta = 2 \cos^2 \theta - 1 \)
Now, the expression becomes:
\( = \tan \theta \cdot \frac{(2 \cos^2 \theta - 1)}{(2 \cos^2 \theta - 1)} \)
Cancel out the common term \( (2 \cos^2 \theta - 1) \):
\( = \tan \theta \cdot 1 \)
\( = \tan \theta \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: First, take out \( \sin \theta \) from the top and \( \cos \theta \) from the bottom. This leaves \( \tan \theta \) and a fraction. Then, use the rule that \( \sin^2 \theta = 1 - \cos^2 \theta \) in the numerator of that fraction. After simplifying, you'll find the top and bottom of that fraction are the same, so they cancel out, leaving just \( \tan \theta \).

Exam Tip: When you see expressions with cubes or higher powers, try factoring out the lowest power first. Then, look for ways to use the Pythagorean identities to simplify the remaining terms, especially those that appear similar in structure (like \( 1-2\sin^2 \theta \) and \( 2\cos^2 \theta - 1 \)).

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(viii) \( (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)
Answer:
Let's start with the Left Hand Side (LHS):
\( \text{LHS} = (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 \)
Expand both squared terms using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( = (\sin^2 A + 2 \sin A \text{cosec } A + \text{cosec}^2 A) + (\cos^2 A + 2 \cos A \sec A + \sec^2 A) \)
We know that \( \sin A \text{cosec } A = \sin A \cdot \frac{1}{\sin A} = 1 \) and \( \cos A \sec A = \cos A \cdot \frac{1}{\cos A} = 1 \).
Substitute these values:
\( = (\sin^2 A + 2 \cdot 1 + \text{cosec}^2 A) + (\cos^2 A + 2 \cdot 1 + \sec^2 A) \)
\( = \sin^2 A + 2 + \text{cosec}^2 A + \cos^2 A + 2 + \sec^2 A \)
Rearrange terms to group \( \sin^2 A + \cos^2 A \):
\( = (\sin^2 A + \cos^2 A) + 2 + 2 + \text{cosec}^2 A + \sec^2 A \)
Using the identity \( \sin^2 A + \cos^2 A = 1 \):
\( = 1 + 4 + \text{cosec}^2 A + \sec^2 A \)
\( = 5 + \text{cosec}^2 A + \sec^2 A \)
Now, use the identities \( \text{cosec}^2 A = 1 + \cot^2 A \) and \( \sec^2 A = 1 + \tan^2 A \):
\( = 5 + (1 + \cot^2 A) + (1 + \tan^2 A) \)
\( = 5 + 1 + \cot^2 A + 1 + \tan^2 A \)
\( = 7 + \tan^2 A + \cot^2 A \)
This is the Right Hand Side (RHS).
Thus, \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: Expand both squared brackets. Remember that \( \sin A \cdot \text{cosec } A = 1 \) and \( \cos A \cdot \sec A = 1 \). Group \( \sin^2 A + \cos^2 A \) (which is 1) and combine the numbers. Finally, change \( \text{cosec}^2 A \) to \( (1+\cot^2 A) \) and \( \sec^2 A \) to \( (1+\tan^2 A) \) to get the desired result.

Exam Tip: When dealing with sums of squares of binomials involving reciprocal functions, expand them first. Remember that \( \sin A \text{cosec } A = 1 \) and \( \cos A \sec A = 1 \), which simplifies terms. Then, apply Pythagorean identities for \( \text{cosec}^2 A \) and \( \sec^2 A \) to introduce cotangent and tangent terms.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(ix) \( (\text{cosec } A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A+\cot A} \)
Answer:
Let's simplify both the Left Hand Side (LHS) and Right Hand Side (RHS) separately.

**Left Hand Side (LHS):**
\( \text{LHS} = (\text{cosec } A - \sin A)(\sec A - \cos A) \)
Convert \( \text{cosec } A \) and \( \sec A \) into \( \sin A \) and \( \cos A \):
\( = \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right) \)
Combine terms inside each parenthesis:
\( = \left(\frac{1 - \sin^2 A}{\sin A}\right) \left(\frac{1 - \cos^2 A}{\cos A}\right) \)
Use the identities \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \):
\( = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} \)
Cancel out common terms (one \( \cos A \) and one \( \sin A \)):
\( = \cos A \sin A \)

**Right Hand Side (RHS):**
\( \text{RHS} = \frac{1}{\tan A+\cot A} \)
Convert \( \tan A \) and \( \cot A \) into \( \sin A \) and \( \cos A \):
\( = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} \)
Combine the terms in the denominator by finding a common denominator \( \sin A \cos A \):
\( = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \)
Use the identity \( \sin^2 A + \cos^2 A = 1 \):
\( = \frac{1}{\frac{1}{\sin A \cos A}} \)
Invert and multiply:
\( = \sin A \cos A \)
Since \( \text{LHS} = \sin A \cos A \) and \( \text{RHS} = \sin A \cos A \), then \( \text{LHS} = \text{RHS} \), and the identity is proven.
In simple words: First, simplify the left side by changing cosec A and sec A to sin and cos. Combine terms, then use \( 1 - \sin^2 A = \cos^2 A \) and \( 1 - \cos^2 A = \sin^2 A \) to simplify further, resulting in \( \sin A \cos A \). Next, simplify the right side by changing tan A and cot A to sin and cos. Add them in the denominator, use \( \sin^2 A + \cos^2 A = 1 \), and then flip the fraction. Both sides will be the same.

Exam Tip: For problems where both sides are complex, working on each side independently until they both simplify to the same expression is often the clearest path to proof. Always convert to sine and cosine as a primary simplification strategy.

Question 5. Prove that the following identities, where the angles involved are acute angles for which the expression are defined.
(x) \( \left[\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right] = \left[\frac{1-\tan A}{1-\cot A}\right]^{2} = \tan^2 A \)
Answer:
We need to prove that the first expression equals \( \tan^2 A \) and the second expression also equals \( \tan^2 A \).

**Part 1: Prove \( \left[\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right] = \tan^2 A \)**
Start with the Left Hand Side (LHS) of this part:
\( \text{LHS} = \frac{1+\tan ^{2} A}{1+\cot ^{2} A} \)
Use the identities \( 1 + \tan^2 A = \sec^2 A \) and \( 1 + \cot^2 A = \text{cosec}^2 A \):
\( = \frac{\sec^2 A}{\text{cosec}^2 A} \)
Convert to \( \sin A \) and \( \cos A \):
\( = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \)
\( = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} \)
\( = \frac{\sin^2 A}{\cos^2 A} \)
\( = \tan^2 A \)
This matches the Right Hand Side (RHS) of this part.

**Part 2: Prove \( \left[\frac{1-\tan A}{1-\cot A}\right]^{2} = \tan^2 A \)**
Start with the Left Hand Side (LHS) of this part:
\( \text{LHS} = \left[\frac{1-\tan A}{1-\cot A}\right]^{2} \)
Convert \( \cot A \) into \( \frac{1}{\tan A} \):
\( = \left[\frac{1-\tan A}{1-\frac{1}{\tan A}}\right]^{2} \)
Simplify the denominator:
\( = \left[\frac{1-\tan A}{\frac{\tan A - 1}{\tan A}}\right]^{2} \)
Multiply the numerator by the reciprocal of the denominator:
\( = \left[(1-\tan A) \cdot \frac{\tan A}{\tan A - 1}\right]^{2} \)
Notice that \( (1-\tan A) = -(\tan A - 1) \). Substitute this:
\( = \left[-(\tan A - 1) \cdot \frac{\tan A}{\tan A - 1}\right]^{2} \)
Cancel out \( (\tan A - 1) \):
\( = [-\tan A]^{2} \)
\( = \tan^2 A \)
This matches the Right Hand Side (RHS) of this part.
Since both parts simplify to \( \tan^2 A \), the identity is proven.
In simple words: For the first part, replace \( (1 + \tan^2 A) \) with \( \sec^2 A \) and \( (1 + \cot^2 A) \) with \( \text{cosec}^2 A \). Then, change these to their sin and cos forms to easily get \( \tan^2 A \). For the second part, replace \( \cot A \) with \( 1/\tan A \). Simplify the fraction inside the square, recognizing that \( (1-\tan A) \) is the negative of \( (\tan A - 1) \). After canceling, square the remaining term to also get \( \tan^2 A \).

Exam Tip: When proving multi-part identities, break them down into separate, manageable proofs. Always simplify expressions to a common base (like sine and cosine or tangent) to make comparisons easier. Be careful with signs when factoring or simplifying terms like \( (a-b) = -(b-a) \).