CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 04

Question. The value of sin 60° ⋅ cos 30° + sin 30° ⋅ cos 60° is
(a) 0
(b) 1
(c) 2
(d) 8
Answer : B

Question. The value of (1 + tan2 θ)(1 – sin θ)(1 + sin θ) =
(a) 0
(b) 1
(c) 2
(d) None of these
Answer : B

Question. Value of cos 0° ⋅ cos 30° ⋅ cos 45° ⋅ cos 60° ⋅ cos 90° is
(a) 0
(b) 1
(c) 2
(d) 9
Answer : A

Question. The value of (sin² θ = 1/1+ tan² θ) =
(a) 0
(b) 1
(c) 2
(d) 5
Answer : B

Question. 2 tan² 45° + cos² 30° – sin² 60° equals
(a) 1
(b) 2
(c) 5
(d) 6
Answer : B

Question. If 15 cot A = 8, then the value of cosec A is
(a) 15/12
(b) 13/15
(c) 4/15
(d) 17/15
Answer : D

Question. If tan(A + B) = √3 and tan(A – B) = 1/√3, A > B, then the value of A is
(a) A = 30°
(b) A = 60°
(c) A = 90°
(d) A = 45°
Answer : D

Question. The value of sin 60° cos 30° + sin 30° cos 60° is
(a) 1
(b) 2
(c) 11
(d) 0
Answer : A

Question. The value of cos30° + sin 60°/1 + cos60° + sin30° is
(a) √3/2
(b) 2/√3
(c) 1/√2
(d) 0
Answer : A

Question. ABC is a triangle right angled at C and AC = v3 BC. Then ∠ABC =
(a) 30°
(b) 60°
(c) 90°
(d) 0°
Answer : B

Question. Evaluate: 4 sin² 60° + 3 tan² 30° – 8 sin 45° cos 45°
(a) 0
(b) 1
(c) 2
(d) 5
Answer : A

Question. If sin θ = x and sec q = y, then the value of cot θ is
(a) xy
(b) 2xy
(c) 1/xy
(d) x + y
Answer : C

Question. If (1 + cos A)(1 – cos A) = 3/4, the value of sec A is
(a) 2
(b) –2
(c) ±2
(d) 0
Answer : C

Question. Evaluate: sin30° + tan45° - cosec60°/sec30° + cos 60° + cot45°
(a) 3√3+2/3√3+2
(b) 3√3-4/3√3+4
(c) 3√3+8/3√3-9
(d) None of these
Answer : B
 

Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose
the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): In a right-angled triangle, if tan θ = 3/4, the greatest side of the triangle is 5 units.
Reason (R): (Greatest side)² = (Hypotenuse)2 = (Perpendicular)2 + (Base)².
Answer : A

Question. Assertion (A): In a right-angled triangle, if cos θ = 1/2 and sin θ = 3/2, then tan θ = 3 .
Reason (R): sinθ/cosθ
Answer : A

ONE MARKERS

1) If θ = 45⁰, the value of cosec² θ is a) 1/√2 b)1 c) 1/2 d) 2

2) sin (60+ θ ) – cos (30 – θ ) is equal to a)2 cos θ b) 2 sin θ c) 0 d) 1

3) If sin A + sin² A=1, then the value of cos²A + cos 4A is

a) 2 b) 1 c) – 2 d) 0

4) The value of [(sec A + tan A) (1 – sin A)] is equal to

a) tan² A b) sin² A c) cos A d) sin A.

5) In fig .if D is mid-point of BC, the value of tan x⁰/ tan y⁰ is A

∟CAD= x⁰, ∟CAB =y⁰

C D B

a) 1/3 b) 1 c) 2 d) 1 / 2

6) If x = 3 sec² θ – 1, y = tan² θ – 2 then x – 3y is equal to

a) 3 b) 4 c) 8 d) 5

7) In the given fig. ∟ACB =90⁰,∟BDC = 90⁰,CD = 4cm, BD = 3 cm, AC =12 cm.cos A – sinA Is equal to a) 5/12 b) 5/13 c) 7/12 d) 7/13. A C

8) The value of cos (90 - θ θ ) cos θ – 1 is DDDD tan θ B

a) − sin² θ b) − cosec² θ c) − cos² θ d) − cot θ

9) If tan θ + 1/ tan θ = 2, then the value of tan² θ + 1/ tan² θ is

a) 3 b) 4 c) 2 d) − 4

10) If sec 4A = cosec (A −20⁰), where 4A is an acute angle, then the value of A is

a) 21⁰ b) 22⁰ c) 23⁰ d) 24⁰.

TWO MARKERS

1) If 7 sin² θ + 3 cos² θ = 4, show that tan θ = 1 / √3.

Question 1) If θ = 45⁰, the value of cosec² θ is a) 1/2 b)1 c) 1/2 d) 2
Answer: (d) 2
We are given \(\theta = 45^\circ\).
Since \(\csc 45^\circ = \sqrt{2}\), we have:
\[ \csc^2 45^\circ = (\sqrt{2})^2 = 2 \]
In simple words: Substituting 45 degrees into cosecant gives the square root of 2. Squaring this value results in 2.

Exam Tip: Memorize the standard values of basic trigonometric ratios for \(30^\circ\), \(45^\circ\), and \(60^\circ\) to solve single-mark questions quickly.

Question 2) sin (60+ θ ) – cos (30 – θ ) is equal to a)2 cos θ b) 2 sin θ c) 0 d) 1
Answer: (c) 0
Using the complementary angle identity \(\cos(90^\circ - x) = \sin x\):
\[ \cos(30^\circ - \theta) = \sin[90^\circ - (30^\circ - \theta)] = \sin(60^\circ + \theta) \]
Substituting this back into the expression:
\[ \sin(60^\circ + \theta) - \sin(60^\circ + \theta) = 0 \]
In simple words: Since sine and cosine are complementary, the sine of an angle is equal to the cosine of its complement. This makes the two subtracted terms identical, so their difference is zero.

Exam Tip: If two angles add up to \(90^\circ\), their respective sine and cosine values are always equal.

Question 3) If sin A + sin² A=1, then the value of cos²A + cos 4 A is a) 2 b) 1 c) – 2 d) 0
Answer: (b) 1
We are given:
\[ \sin A + \sin^2 A = 1 \implies \sin A = 1 - \sin^2 A = \cos^2 A \]
Now, we evaluate the expression:
\[ \cos^2 A + \cos^4 A = \cos^2 A + (\cos^2 A)^2 \]
Substituting \(\cos^2 A = \sin A\):
\[ = \sin A + \sin^2 A = 1 \]
In simple words: Use the Pythagorean identity to rewrite the given equation as sine equals cosine squared. Substituting this relation into the required expression simplifies it back to the original equation, which equals 1.

Exam Tip: When given a trigonometric equation equal to 1, try isolating one term on one side to find useful substitutions for higher-power terms.

Question 4) The value of [(sec A + tan A) (1 – sin A)] is equal to a) tan² A b) sin² A c) cos A d) sin A.
Answer: (c) cos A
Express \(\sec A\) and \(\tan A\) in terms of sine and cosine:
\[ \sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} \]
Now, multiply this by \((1 - \sin A)\):
\[ (\sec A + \tan A)(1 - \sin A) = \left(\frac{1 + \sin A}{\cos A}\right)(1 - \sin A) = \frac{1 - \sin^2 A}{\cos A} \]
Using the identity \(1 - \sin^2 A = \cos^2 A\):
\[ = \frac{\cos^2 A}{\cos A} = \cos A \]
In simple words: Convert the secant and tangent terms into sines and cosines. Multiplying them out creates a numerator equal to cosine squared, which cancels with the denominator to leave cosine.

Exam Tip: Expressing complex trigonometric functions in terms of sine and cosine is a highly reliable strategy to simplify products and fractions.

Question 5) In fig .if D is mid-point of BC, the value of tan x⁰/ tan y⁰ is lCAD= x⁰, lCAB =y⁰ a) 1/3 b) 1 c) 2 d) 1 / 2
Answer: (d) 1 / 2
In the right-angled triangle \(AC\delta\) at \(C\), where \(D\) is on side \(BC\):
\[ \tan x^\circ = \frac{CD}{AC} \]
In right-angled triangle \(ACB\) at \(C\):
\[ \tan y^\circ = \frac{BC}{AC} \]
Now, evaluate the ratio:
\[ \frac{\tan x^\circ}{\tan y^\circ} = \frac{\frac{CD}{AC}}{\frac{BC}{AC}} = \frac{CD}{BC} \]
Since \(D\) is the midpoint of \(BC\), we have \(CD = \frac{1}{2} BC\). Therefore:
\[ \frac{\tan x^\circ}{\tan y^\circ} = \frac{\frac{1}{2} BC}{BC} = \frac{1}{2} \]
In simple words: Find the tangent of both angles using the side lengths. The ratio of the tangents depends on the ratio of the opposite sides, which is one-half because \(D\) is the midpoint.

Exam Tip: For problems involving triangles with a common side, expressing trigonometric ratios in terms of that common side allows easy cancellation.

Question 6) If x = 3 sec² θ – 1, y = tan² θ – 2 then x – 3y is equal to a) 3 b) 4 c) 8 d) 5
Answer: (c) 8
Substitute the expressions for \(x\) and \(y\) into \(x - 3y\):
\[ x - 3y = (3 \sec^2 \theta - 1) - 3(\tan^2 \theta - 2) \]
\[ = 3 \sec^2 \theta - 1 - 3 \tan^2 \theta + 6 \]
\[ = 3(\sec^2 \theta - \tan^2 \theta) + 5 \]
Since \(\sec^2 \theta - \tan^2 \theta = 1\):
\[ = 3(1) + 5 = 8 \]
In simple words: Group the secant and tangent terms to use the Pythagorean identity. This simplifies the variable terms to a constant, leaving a final answer of 8.

Exam Tip: Always look to group terms of the same angle that form standard Pythagorean identities to eliminate trigonometric variables.

Question 7) In the given fig. ∟ACB =90⁰,∟BDC = 90⁰,CD = 4cm, BD = 3 cm, AC =12 cm.cos A – sinA Is equal to a) 5/12 b) 5/13 c) 7/12 d) 7/13.
Answer: (d) 7/13
In right-angled triangle \(\Delta BDC\), using Pythagoras theorem:
\[ BC = \sqrt{CD^2 + BD^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\text{ cm} \]
In right-angled triangle \(\Delta ACB\):
\[ AB = \sqrt{AC^2 + BC^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ cm} \]
Now, find \(\cos A\) and \(\sin A\) from right-angled triangle \(\Delta ACB\):
\[ \cos A = \frac{AC}{AB} = \frac{12}{13} \]
\[ \sin A = \frac{BC}{AB} = \frac{5}{13} \]
Subtracting these two values:
\[ \cos A - \sin A = \frac{12}{13} - \frac{5}{13} = \frac{7}{13} \]
In simple words: Find the side lengths of the right triangles using Pythagoras theorem. Use these side lengths to write the sine and cosine fractions and perform the subtraction.

Exam Tip: Be careful to identify which angle is being used to determine which side is the perpendicular (opposite) and which is the base (adjacent).

Question 8) The value of cos (90 - θ θ ) cos θ – 1 is tan θ a) − sin² θ b) − cosec² θ c) − cos² θ d) − cot θ
Answer: (a) − sin² θ
The given expression is:
\[ \frac{\cos(90^\circ - \theta)\cos\theta}{\tan\theta} - 1 \]
Since \(\cos(90^\circ - \theta) = \sin\theta\) and \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):
\[ = \frac{\sin\theta\cos\theta}{\frac{\sin\theta}{\cos\theta}} - 1 = \cos^2\theta - 1 \]
Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\[ \cos^2\theta - 1 = -\sin^2\theta \]
In simple words: Convert the complementary term and tangent. Simplifying the fraction leaves cosine squared, which can be rewritten as negative sine squared.

Exam Tip: Remember that \(1 - \cos^2\theta = \sin^2\theta\), which means \(\cos^2\theta - 1 = -\sin^2\theta\). Pay close attention to signs.

Question 9) If tan θ + 1/ tan θ = 2, then the value of tan² θ + 1/ tan² θ is a) 3 b) 4 c) 2 d) − 4
Answer: (c) 2
We are given:
\[ \tan\theta + \frac{1}{\tan\theta} = 2 \]
Squaring both sides:
\[ \left(\tan\theta + \frac{1}{\tan\theta}\right)^2 = 2^2 \]
\[ \tan^2\theta + \frac{1}{\tan^2\theta} + 2\tan\theta\left(\frac{1}{\tan\theta}\right) = 4 \]
\[ \tan^2\theta + \frac{1}{\tan^2\theta} + 2 = 4 \]
\[ \tan^2\theta + \frac{1}{\tan^2\theta} = 2 \]
In simple words: Square both sides of the given equation. The middle term of the expanded square simplifies to 2, which you subtract from 4 to find the answer.

Exam Tip: Squaring equations of the form \(x + \frac{1}{x} = k\) always yields \(x^2 + \frac{1}{x^2} = k^2 - 2\).

Question 10) If sec 4A = cosec (A −20⁰), where 4A is an acute angle, then the value of A is a) 21⁰ b) 22⁰ c) 23⁰ d) 24⁰.
Answer: (b) 22⁰
Using the identity \(\sec x = \csc(90^\circ - x)\):
\[ \csc(90^\circ - 4A) = \csc(A - 20^\circ) \]
Equating the angles:
\[ 90^\circ - 4A = A - 20^\circ \]
\[ 5A = 110^\circ \implies A = 22^\circ \]
In simple words: Convert the secant term to cosecant of its complement. This lets you compare the two angles directly as a linear equation to find \(A\).

Exam Tip: Verify that the resulting value of \(A\) makes \(4A\) an acute angle (\(4 \times 22^\circ = 88^\circ < 90^\circ\)) as stated in the question.

Question 1) If 7 sin² θ + 3 cos² θ = 4, show that tan θ = 1 / √3.
Answer:
Divide the entire equation by \(\cos^2\theta\):
\[ \frac{7\sin^2\theta}{\cos^2\theta} + \frac{3\cos^2\theta}{\cos^2\theta} = \frac{4}{\cos^2\theta} \]
\[ 7\tan^2\theta + 3 = 4\sec^2\theta \]
Since \(\sec^2\theta = 1 + \tan^2\theta\):
\[ 7\tan^2\theta + 3 = 4(1 + \tan^2\theta) \]
\[ 7\tan^2\theta + 3 = 4 + 4\tan^2\theta \]
\[ 3\tan^2\theta = 1 \implies \tan^2\theta = \frac{1}{3} \]
Taking the positive square root for an acute angle \(\theta\):
\[ \tan\theta = \frac{1}{\sqrt{3}} \]
Hence proved.
In simple words: Divide the equation by cosine squared to convert all terms to tangent and secant. Replacing secant squared with one plus tangent squared lets you isolate and solve for tangent.

Exam Tip: Dividing equations of sine and cosine by \(\cos^2\theta\) is a standard technique to transform them into easily solvable equations of \(\tan\theta\).

Question 2) If sec α = 5 / 4, evaluate 1 – tan α / 1 + tan α .
Answer:
Since \(\sec\alpha = \frac{5}{4} = \frac{\text{Hypotenuse}}{\text{Base}}\).
Using Pythagoras theorem, we find the perpendicular:
\[ \text{Perpendicular} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3 \]
Thus:
\[ \tan\alpha = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4} \]
Now, evaluate the given expression:
\[ \frac{1 - \tan\alpha}{1 + \tan\alpha} = \frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} = \frac{\frac{1}{4}}{\frac{7}{4}} = \frac{1}{7} \]
In simple words: Find the missing perpendicular side of the right triangle using Pythagoras theorem. Use this to write tangent as a fraction and substitute it into the given expression.

Exam Tip: When evaluating fractions with rational sub-fractions, simplify the numerator and denominator separately before dividing.

Question 3) If A and B are acute angles such that tan A = 1/ 2 and tan B =1/3 and Tan (A+B) = tan A+ tan B, find A+B / 1− tan A tan B
Answer:
Using the given addition formula:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} \]
Substitute the values of \(\tan A\) and \(\tan B\):
\[ \tan(A+B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2}\right)\left(\frac{1}{3}\right)} = \frac{\frac{5}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1 \]
Since \(\tan(A+B) = 1\), and \(A\) and \(B\) are acute:
\[ A+B = 45^\circ \]
In simple words: Substitute the fractions into the given formula. This simplifies the right side to 1, which corresponds to an angle sum of 45 degrees.

Exam Tip: Be comfortable with fraction operations in both the numerator and denominator to avoid algebraic mistakes.

Question 4) Prove that sin A + cos A / Sin A – cos A + sinA − cos A / sin A + cosA = 2 / sin²A − cos²A .
Answer:
Let us take the Left-Hand Side (LHS) and combine the fractions using a common denominator:
\[ \text{LHS} = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A - \cos A)(\sin A + \cos A)} \]
Expand the squares in the numerator:
\[ = \frac{(\sin^2 A + \cos^2 A + 2\sin A\cos A) + (\sin^2 A + \cos^2 A - 2\sin A\cos A)}{\sin^2 A - \cos^2 A} \]
The cross-product terms cancel out:
\[ = \frac{2(\sin^2 A + \cos^2 A)}{\sin^2 A - \cos^2 A} \]
Since \(\sin^2 A + \cos^2 A = 1\):
\[ = \frac{2}{\sin^2 A - \cos^2 A} = \text{RHS} \]
Hence proved.
In simple words: Find a common denominator to add the two fractions. The middle terms in the expanded numerator cancel out, leaving twice the sum of sine squared and cosine squared, which simplifies to 2.

Exam Tip: Grouping terms to utilize the identity \(\sin^2 A + \cos^2 A = 1\) is a very powerful way to simplify long proofs.

Question 5) Find acute angles A and B ,if sin (A+2B) = √3 /2 and cos (A+4B) = 0
Answer:
We are given:
\[ \sin(A+2B) = \frac{\sqrt{3}}{2} \implies A + 2B = 60^\circ \quad \text{--- (Equation 1)} \]
\[ \cos(A+4B) = 0 \implies A + 4B = 90^\circ \quad \text{--- (Equation 2)} \]
Subtracting Equation 1 from Equation 2:
\[ (A + 4B) - (A + 2B) = 90^\circ - 60^\circ \]
\[ 2B = 30^\circ \implies B = 15^\circ \]
Substituting \(B = 15^\circ\) into Equation 1:
\[ A + 2(15^\circ) = 60^\circ \implies A = 30^\circ \]
Thus, \(A = 30^\circ\) and \(B = 15^\circ\).
In simple words: Use standard angles to write two linear equations from the sine and cosine values. Solving these simultaneous equations yields the values for \(A\) and \(B\).

Exam Tip: Elimination is the most straightforward method to solve the resulting linear system once you have removed the trigonometric functions.

Question 6) Find the value of tan60⁰,geometrically.
Answer:
Consider an equilateral triangle \(ABC\) with side length \(2a\). Each angle in this triangle is \(60^\circ\).
Draw a perpendicular altitude \(AD\) from vertex \(A\) to side \(BC\).
Since \(AD\) is the altitude, it bisects \(BC\), so \(BD = DC = a\), and \(\angle BAD = 30^\circ\).
In right-angled triangle \(ABD\), using Pythagoras theorem:
\[ AB^2 = AD^2 + BD^2 \]
\[ (2a)^2 = AD^2 + a^2 \]
\[ 4a^2 = AD^2 + a^2 \implies AD^2 = 3a^2 \]
\[ AD = a\sqrt{3} \]
Now, in right-angled triangle \(ABD\), for angle \(\angle B = 60^\circ\):
\[ \tan 60^\circ = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AD}{BD} = \frac{a\sqrt{3}}{a} = \sqrt{3} \]
Thus, \(\tan 60^\circ = \sqrt{3}\).
In simple words: Draw an equilateral triangle and cut it in half with an altitude line. Applying Pythagoras theorem to the resulting right triangle allows you to find the ratio of the height to the base, which is tangent of 60 degrees.

Exam Tip: A neat, labeled geometric diagram is essential to secure full marks in geometric derivations.

Question 7) Without using trigonometric tables,evaluate: 11 sin 70⁰ / 7 cos 20⁰ - 4 cos 53⁰ cosec 37⁰ / 7 tan15⁰tan35⁰tan55⁰tan75
Answer:
Let's simplify the first term:
Since \(\sin 70^\circ = \cos(90^\circ - 70^\circ) = \cos 20^\circ\):
\[ \frac{11}{7} \cdot \frac{\sin 70^\circ}{\cos 20^\circ} = \frac{11}{7} \cdot \frac{\cos 20^\circ}{\cos 20^\circ} = \frac{11}{7} \]
Now, let's simplify the second term:
In the numerator, since \(\csc 37^\circ = \frac{1}{\sin 37^\circ}\) and \(\sin 37^\circ = \cos(90^\circ - 37^\circ) = \cos 53^\circ\):
\[ \cos 53^\circ \csc 37^\circ = \frac{\cos 53^\circ}{\sin 37^\circ} = \frac{\cos 53^\circ}{\cos 53^\circ} = 1 \]
In the denominator, group the complementary angles:
\[ \tan 15^\circ \tan 35^\circ \tan 55^\circ \tan 75^\circ = (\tan 15^\circ \tan 75^\circ)(\tan 35^\circ \tan 55^\circ) \]
Since \(\tan 75^\circ = \cot 15^\circ\) and \(\tan 55^\circ = \cot 35^\circ\):
\[ = (\tan 15^\circ \cot 15^\circ)(\tan 35^\circ \cot 35^\circ) = 1 \cdot 1 = 1 \]
So, the second term is:
\[ \frac{4}{7} \cdot \frac{1}{1} = \frac{4}{7} \]
Subtracting the two simplified terms:
\[ \frac{11}{7} - \frac{4}{7} = \frac{7}{7} = 1 \]
In simple words: Convert complementary angles to common ones. This simplifies the fractions to 1, leaving a basic subtraction of fractions that results in 1.

Exam Tip: For product terms of tangent, group angles that sum up to \(90^\circ\) so they convert to cotangent and cancel to 1.

Question 8) In a right triangle ABC, right angled at B, if tan A =1,then verify that 2 sin A cos A =1
Answer:
We are given \(\tan A = 1\), which means \(A = 45^\circ\).
Substitute this value of \(A\) into the expression:
\[ 2\sin A\cos A = 2\sin 45^\circ\cos 45^\circ \]
We know \(\sin 45^\circ = \frac{1}{\sqrt{2}}\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\):
\[ = 2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) = 2\left(\frac{1}{2}\right) = 1 \]
Hence verified.
In simple words: Find the angle whose tangent is 1, which is 45 degrees. Plug this angle into the expression to verify that it multiplies to 1.

Exam Tip: Identifying the angle value directly from simple ratios is the easiest way to perform verification proofs.

Question 9) Prove that( 1+cot² θ )(1 – cos θ)(1 + cos θ) = 1.
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = (1 + \cot^2\theta)(1 - \cos\theta)(1 + \cos\theta) \]
Using the identity \(1 + \cot^2\theta = \csc^2\theta\):
\[ = \csc^2\theta \cdot (1 - \cos^2\theta) \]
Using the identity \(1 - \cos^2\theta = \sin^2\theta\):
\[ = \csc^2\theta \cdot \sin^2\theta \]
Since cosecant and sine are reciprocals, their product is 1:
\[ = 1 = \text{RHS} \]
Hence proved.
In simple words: Rewrite the first bracket as cosecant squared, and multiply the last two brackets to get sine squared. Multiplying these two reciprocals results in 1.

Exam Tip: Be comfortable with binomial expansions of the form \((1 - x)(1 + x) = 1 - x^2\) to simplify groups of terms quickly.

Question 10) If sin θ + cos θ = √2 sin (90 – θ ), determine cos θ.
Answer:
First, we use the complementary identity \(\sin(90^\circ - \theta) = \cos\theta\) to rewrite the equation:
\[ \sin\theta + \cos\theta = \sqrt{2}\cos\theta \]
Subtract \(\cos\theta\) from both sides:
\[ \sin\theta = \sqrt{2}\cos\theta - \cos\theta = (\sqrt{2} - 1)\cos\theta \]
Now, we find \(\cot\theta\):
\[ \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{1}{\sqrt{2} - 1} \]
Rationalize the denominator:
\[ \cot\theta = \frac{1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 \]
In simple words: Convert the complementary sine term on the right. Group the cosine terms to find the ratio of cosine over sine, which is cotangent, and rationalize the answer.

Exam Tip: Note that "determine cos \(\theta\)" in the original text is a common typographical error for "determine cot \(\theta\)", as indicated by the matching answer in the key.

Question 1) Prove that tan θ – cot θ = tan² θ − cot² θ / Sin θ cos θ
Answer:
Let's simplify the Left-Hand Side (LHS):
\[ \text{LHS} = \frac{\tan\theta - \cot\theta}{\sin\theta\cos\theta} = \frac{\tan\theta}{\sin\theta\cos\theta} - \frac{\cot\theta}{\sin\theta\cos\theta} \delta \]
Substitute \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and \(\cot\theta = \frac{\cos\theta}{\sin\theta}\):
\[ = \frac{\frac{\sin\theta}{\cos\theta}}{\sin\theta\cos\theta} - \frac{\frac{\cos\theta}{\sin\theta}}{\sin\theta\cos\theta} \]
\[ = \frac{1}{\cos^2\theta} - \frac{1}{\sin^2\theta} = \sec^2\theta - \csc^2\theta \]
Now substitute \(\sec^2\theta = 1 + \tan^2\theta\) and \(\csc^2\theta = 1 + \cot^2\theta\):
\[ = (1 + \tan^2\theta) - (1 + \cot^2\theta) = \tan^2\theta - \cot^2\theta = \text{RHS} \]
Hence proved.
In simple words: Split the fraction into two parts and express everything in terms of sine and cosine. Simplifying the fractions gives secant and cosecant squared, which convert to the tangent and cotangent squares.

Exam Tip: Splitting fractions with multiple terms in the numerator is often a very useful step to simplify expressions.

Question 2) If cos θ + sin θ = √2cos θ, show that cos θ – sin θ =√2 sin θ.
Answer:
We are given:
\[ \cos\theta + \sin\theta = \sqrt{2}\cos\theta \]
Subtract \(\cos\theta\) from both sides:
\[ \sin\theta = \sqrt{2}\cos\theta - \cos\theta = (\sqrt{2} - 1)\cos\theta \]
Multiply both sides by \((\sqrt{2} + 1)\):
\[ (\sqrt{2} + 1)\sin\theta = (\sqrt{2} + 1)(\sqrt{2} - 1)\cos\theta \]
\[ \sqrt{2}\sin\theta + \sin\theta = (2 - 1)\cos\theta = \cos\theta \]
Rearranging the terms:
\[ \cos\theta - \sin\theta = \sqrt{2}\sin\theta \]
Hence proved.
In simple words: Group the cosine terms and isolate sine. Multiplying by the conjugate of the coefficient rationalizes the expression and directly leads to the desired formula.

Exam Tip: This is a very common proof. Learn the algebraic step of multiplying by the conjugate to reverse the coefficients of sine and cosine.

Question 3) Prove that sin θ = 2 + sin θ / cot θ + cosec θ cot θ − cosec θ.
Answer:
Let's rearrange the target equation to show:
\[ \frac{\sin\theta}{\cot\theta + \csc\theta} - \frac{\sin\theta}{\cot\theta - \csc\theta} = 2 \]
Let's evaluate the Left-Hand Side (LHS) of this rearranged equation:
\[ \text{LHS} = \sin\theta \left( \frac{1}{\cot\theta + \csc\theta} - \frac{1}{\cot\theta - \csc\theta} \right) \]
Combine the terms using a common denominator:
\[ = \sin\theta \left( \frac{(\cot\theta - \csc\theta) - (\cot\theta + \csc\theta)}{\cot^2\theta - \csc^2\theta} \right) \]
\[ = \sin\theta \left( \frac{-2\csc\theta}{\cot^2\theta - \csc^2\theta} \right) \]
Since \(\cot^2\theta - \csc^2\theta = -1\):
\[ = \sin\theta \left( \frac{-2\csc\theta}{-1} \right) = \sin\theta \cdot 2\csc\theta \]
Since \(\csc\theta = \frac{1}{\sin\theta}\):
\[ = \sin\theta \cdot \frac{2}{\sin\theta} = 2 = \text{RHS} \]
Hence proved.
In simple words: Move the fractions to the same side and combine them using cross-multiplication. The denominator simplifies to negative one using a standard identity, which cancels with the numerator to give 2.

Exam Tip: Be careful with the identity \(\csc^2\theta - \cot^2\theta = 1\), which means \(\cot^2\theta - \csc^2\theta = -1\). Do not drop the negative sign.

Question 4) Prove the following identity : (cosec A – sin A)(sec A− cos A)(tanA +cot A) = 1.
Answer:
Let's convert each term of the Left-Hand Side (LHS) into sines and cosines:
\[ \text{LHS} = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right)\left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right) \]
\[ = \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right)\left(\frac{\sin^2 A + \cos^2 A}{\sin A\cos A}\right) \delta \]
Using standard Pythagorean identities:
\[ = \left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right)\left(\frac{1}{\sin A\cos A}\right) \]
\[ = \frac{\sin^2 A\cos^2 A}{\sin^2 A\cos^2 A} = 1 = \text{RHS} \]
Hence proved.
In simple words: Change all functions to sine and cosine. Simplifying the fractions yields terms that cancel each other out completely, resulting in 1.

Exam Tip: Simplify each parenthesis individually before trying to multiply them out.

Question 5) If x/a cos θ + y/b sin θ =1 and x/a sin θ – y/b cos θ = − 1,prove that x²/a²/ +y²/b² =2.
Answer:
Let \(X = \frac{x}{a}\) and \(Y = \frac{y}{b}\).
The equations are:
1. \(X\cos\theta + Y\sin\theta = 1\)
2. \(X\sin\theta - Y\cos\theta = -1\)
Squaring and adding both equations:
\[ (X\cos\theta + Y\sin\theta)^2 + (X\sin\theta - Y\cos\theta)^2 = 1^2 + (-1)^2 \]
\[ (X^2\cos^2\theta + Y^2\sin^2\theta + 2XY\sin\theta\cos\theta) + (X^2\sin^2\theta + Y^2\cos^2\theta - 2XY\sin\theta\cos\theta) = 2 \]
The cross-product terms cancel out:
\[ X^2(\cos^2\theta + \sin^2\theta) + Y^2(\sin^2\theta + \cos^2\theta) = 2 \]
Since \(\sin^2\theta + \cos^2\theta = 1\):
\[ X^2 + Y^2 = 2 \implies \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \]
Hence proved.
In simple words: Square both equations. When you add them together, the cross-product terms cancel, and grouping the rest simplifies to twice the standard Pythagorean identity, which equals 2.

Exam Tip: Squaring and adding is the most standard algebraic technique to eliminate trigonometric variables when they are multiplied by sine and cosine.

Question 1) If 2 cos θ –sin θ = x and cos θ – 3 sin θ =y. Prove that 2x² + y² −2xy =5.
Answer:
Let's rewrite the target expression algebraically:
\[ 2x^2 + y^2 - 2xy = x^2 + (x - y)^2 \]
Let's calculate \(x - y\):
\[ x - y = (2\cos\theta - \sin\theta) - (\cos\theta - 3\sin\theta) = \cos\theta + 2\sin\theta \]
Now, substitute the expressions for \(x\) and \(x-y\):
\[ x^2 + (x - y)^2 = (2\cos\theta - \sin\theta)^2 + (\cos\theta + 2\sin\theta)^2 \]
\[ = (4\cos^2\theta + \sin^2\theta - 4\sin\theta\cos\theta) + (\cos^2\theta + 4\sin^2\theta + 4\sin\theta\cos\theta) \]
The cross-product terms cancel out:
\[ = 5\cos^2\theta + 5\sin^2\theta = 5(\cos^2\theta + \sin^2\theta) = 5(1) = 5 \]
Hence proved.
In simple words: Rewrite the target algebraic expression to make the grouping simpler. Substituting the expressions for the variables simplifies the calculations and results in 5.

Exam Tip: Look for algebraic factoring simplifications in the target expression before substituting trigonometric variables to save time.

Question 2) If tan θ + sin θ =m ,and tan θ – sin θ = n,show that (m² − n²)² = 16 mn.
Answer:
Let's calculate the Left-Hand Side (LHS) of the equation:
\[ m^2 - n^2 = (m + n)(m - n) \]
Here:
- \(m + n = (\tan\theta + \sin\theta) + (\tan\theta - \sin\theta) = 2\tan\theta\)
- \(m - n = (\tan\theta + \sin\theta) - (\tan\theta - \sin\theta) = 2\sin\theta\)
So:
\[ m^2 - n^2 = (2\tan\theta)(2\sin\theta) = 4\tan\theta\sin\theta \]
Squaring both sides:
\[ \text{LHS} = (m^2 - n^2)^2 = 16\tan^2\theta\sin^2\theta \]
Now, let's calculate the Right-Hand Side (RHS) of the equation:
\[ mn = (\tan\theta + \sin\theta)(\tan\theta - \sin\theta) = \tan^2\theta - \sin^2\theta \]
\[ = \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta = \sin^2\theta\left(\frac{1}{\cos^2\theta} - 1\right) = \sin^2\theta(\sec^2\theta - 1) = \sin^2\theta\tan^2\theta \]
Multiply by 16:
\[ \text{RHS} = 16mn = 16\tan^2\theta\sin^2\theta \]
Since \(\text{LHS} = \text{RHS}\), the relation is proved.
In simple words: Use the difference of squares on the left side to get a product of tangent and sine. Simplifying the right side product shows both sides are identical.

Exam Tip: Proving that \(\tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta\) is the core identity step required to solve this problem.

Question 3) If sec θ + tan θ = p ,prove that sin θ = p² − 1 / p² + 1
Answer:
We are given:
\[ \sec\theta + \tan\theta = p \]
Using the identity \(\sec^2\theta - \tan^2\theta = 1 \implies (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1\):
\[ \sec\theta - \tan\theta = \frac{1}{p} \]
Adding the two equations:
\[ 2\sec\theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec\theta = \frac{p^2 + 1}{2p} \implies \cos\theta = \frac{2p}{p^2 + 1} \]
Subtracting the second equation from the first:
\[ 2\tan\theta = p - \frac{1}{p} = \frac{p^2 - 1}{p} \implies \tan\theta = \frac{p^2 - 1}{2p} \]
Now, we calculate \(\sin\theta\):
\[ \sin\theta = \tan\theta \cdot \cos\theta = \left(\frac{p^2 - 1}{2p}\right)\left(\frac{2p}{p^2 + 1}\right) = \frac{p^2 - 1}{p^2 + 1} \]
Hence proved.
In simple words: Set up two equations using the secant-tangent identity. Solve for secant and tangent separately, and multiply the tangent by the reciprocal of secant (cosine) to find the sine formula.

Exam Tip: If \(\sec\theta + \tan\theta = p\), then \(\sec\theta - \tan\theta\) is always the reciprocal, \(1/p\). This is a very useful property to remember.

Question 4) If x = tan A+ sin A and y = tan A – sin A , show that x² − y² = 4 √xy.
Answer:
Let's simplify the Left-Hand Side (LHS) of the equation:
\[ \text{LHS} = x^2 - y^2 = (x+y)(x-y) \]
Substitute \(x\) and \(y\):
- \(x+y = 2\tan A\)
- \(x-y = 2\sin A\)
So, \(x^2 - y^2 = 4\tan A\sin A\).
Now, let's simplify the Right-Hand Side (RHS):
\[ \text{RHS} = 4\sqrt{xy} = 4\sqrt{(\tan A + \sin A)(\tan A - \sin A)} = 4\sqrt{\tan^2 A - \sin^2 A} \]
\[ = 4\sqrt{\frac{\sin^2 A}{\cos^2 A} - \sin^2 A} = 4\sqrt{\sin^2 A \left(\frac{1}{\cos^2 A} - 1\right)} \]
\[ = 4\sqrt{\sin^2 A(\sec^2 A - 1)} = 4\sqrt{\sin^2 A\tan^2 A} = 4\tan A\sin A \]
Since \(\text{LHS} = \text{RHS}\), the relation is proved.
In simple words: Expand both sides independently. The left-hand side simplifies to a product using the difference of squares, while the right-hand side simplifies to the same product by factoring out sine under the square root.

Exam Tip: This question is algebraically identical to Four Markers Q2. Memorize this pattern as it frequently appears in examinations.

Question 5) If sec θ = x +1/x, prove that sec θ + tan θ = 2x or 1/2x.
Answer:
Note: The original question contains a standard textbook typo. The correct equation to prove is: if \(\sec\theta = x + \frac{1}{4x}\), then \(\sec\theta + \tan\theta = 2x\) or \(\frac{1}{2x}\). Let us solve this corrected formulation.
We know the identity:
\[ \tan^2\theta = \sec^2\theta - 1 \]
Substitute the value of \(\sec\theta\):
\[ \tan^2\theta = \left(x + \frac{1}{4x}\right)^2 - 1 = x^2 + \frac{1}{16x^2} + 2(x)\left(\frac{1}{4x}\right) - 1 \]
\[ = x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1 = x^2 + \frac{1}{16x^2} - \frac{1}{2} \]
This forms a perfect square:
\[ \tan^2\theta = \left(x - \frac{1}{4x}\right)^2 \implies \tan\theta = \pm \left(x - \frac{1}{4x}\right) \]
Case 1: Taking the positive value:
\[ \sec\theta + \tan\theta = \left(x + \frac{1}{4x}\right) + \left(x - \frac{1}{4x}\right) = 2x \]
Case 2: Taking the negative value:
\[ \sec\theta + \tan\theta = \left(x + \frac{1}{4x}\right) - \left(x - \frac{1}{4x}\right) = \frac{2}{4x} = \frac{1}{2x} \]
Hence proved.
In simple words: Use the identity to express tangent squared in terms of the given secant formula. Squaring creates a perfect square that gives two possible values for tangent, leading to the two requested solutions.

Exam Tip: Be sure to consider both the positive and negative square roots when finding the value of \(\tan\theta\) from \(\tan^2\theta\) to show both parts of the proof.