CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set E

Points to Remember

• If triangle \( ABC \) is right-angled at \( B \) and \( \angle BAC = \theta \), then with reference to the angle \( \theta \), we have
  Base \( (B) = AB \), Perpendicular \( (P) = BC \) and Hypotenuse \( (H) = AC \)
  Also,
  \( \sin \theta = \frac{P}{H} \); \( \cos \theta = \frac{B}{H} \); \( \tan \theta = \frac{P}{B} \); \( \text{cosec } \theta = \frac{H}{P} \);
  \( \sec \theta = \frac{H}{B} \); \( \cot \theta = \frac{B}{P} \).
• \( \sin \theta = \frac{1}{\text{cosec } \theta} \); \( \cos \theta = \frac{1}{\sec \theta} \); \( \tan \theta = \frac{1}{\cot \theta} \)
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
• If \( \theta \) is an acute angle, then
  \( \sin (90^\circ – \theta) = \cos \theta \); \( \cos (90^\circ – \theta) = \sin \theta \); \( \tan (90^\circ – \theta) = \cot \theta \); \( \cot (90^\circ – \theta) = \tan \theta \); \( \sec (90^\circ – \theta) = \text{cosec } \theta \); \( \text{cosec } (90^\circ – \theta) = \sec \theta \).
• \( \sin^2 \theta + \cos^2 \theta = 1 \)
  \( \sec^2 \theta = 1 + \tan^2 \theta \)
  \( \text{cosec}^2 \theta = 1 + \cot^2 \theta \)

Multiple Choice Questions

Question. The value of \( (\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ) \) is equal to ......... 
Answer: 1
Explanation :
\( \tan 1^\circ \tan 2^\circ \dots \tan 89^\circ \)
\( = \tan 1^\circ \tan 2^\circ \dots \tan 45^\circ \dots \tan 88^\circ \tan 89^\circ \)
\( = \tan 1^\circ \tan 2^\circ \dots \tan 45^\circ \dots \tan (90^\circ – 2^\circ) \tan (90^\circ – 1^\circ) \)
\( = \tan 1^\circ \tan 2^\circ \dots \tan 45^\circ \dots \cot 2^\circ \cot 1^\circ \)
[\( \because \tan (90^\circ – \theta) = \cot \theta \)]
\( = \tan 1^\circ \tan 2^\circ \dots \tan 45^\circ \dots \frac{1}{\tan 2^\circ} \frac{1}{\tan 1^\circ} \)
\( = \tan 45^\circ \)
\( = 1 \)

Question. \( \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} \) is equal to :
(a) \( \sin 60^\circ \)
(b) \( \cos 60^\circ \)
(c) \( \tan 60^\circ \)
(d) \( \sin 30^\circ \)
Answer: (a)
Sol. \( \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2 \times 3}{\sqrt{3} \times 4} = \frac{\sqrt{3}}{2} = \sin 60^\circ \).
So, the correct option is (a).

Question. \( 9 \sec^2 A – 9 \tan^2 A \) is equal to :
(a) 1
(b) 9
(c) 8
(d) 0
Answer: (b)
Sol. \( 9 \sec^2 A – 9 \tan^2 A = 9(\sec^2 A – \tan^2 A) = 9(1) = 9 \).
So the correct option is (b).

Question. \( (1 + \tan \theta + \sec \theta) (1 + \cot \theta – \text{cosec } \theta) \) is equal to :
(a) 0
(b) 1
(c) 2
(d) – 1
Answer: (c)
Sol. \( (1 + \tan \theta + \sec \theta) (1 + \cot \theta – \text{cosec } \theta) \)
\( = [1 + \tan \theta + \sec \theta + \cot \theta + \tan \theta \cot \theta + \sec \theta \cot \theta – \text{cosec } \theta – \tan \theta \text{cosec } \theta – \sec \theta \text{cosec } \theta] \)
\( = [1 + \tan \theta + \sec \theta + \cot \theta + 1 + \text{cosec } \theta – \text{cosec } \theta – \sec \theta – \sec \theta \text{cosec } \theta] \)
\( = [2 + \tan \theta + \cot \theta – \sec \theta \text{cosec } \theta] \)
\( = 2 + \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} – \frac{1}{\sin \theta \cos \theta} \)
\( = 2 + \frac{\sin^2 \theta + \cos^2 \theta - 1}{\sin \theta \cos \theta} \)
\( = 2 + \frac{1 - 1}{\sin \theta \cos \theta} = 2 \).
So, the correct option is (c).

Question. \( (\sec A + \tan A) (1 – \sin A) \) is equal to :
(a) \( \sec A \)
(b) \( \sin A \)
(c) \( \text{cosec } A \)
(d) \( \cos A \)
Answer: (d)
Sol. \( (\sec A + \tan A) (1 – \sin A) \)
\( = \left( \frac{1}{\cos A} + \frac{\sin A}{\cos A} \right) (1 - \sin A) \)
\( = \left( \frac{1 + \sin A}{\cos A} \right) (1 - \sin A) \)
\( = \frac{1 - \sin^2 A}{\cos A} \)
\( = \frac{\cos^2 A}{\cos A} = \cos A \).
So, the correct option is (d).

Question. If \( 3 \sin \theta = 2 \), then \( \text{cosec } \theta \) is equal to :
(a) \( \frac{2}{3} \)
(b) \( \frac{3}{2} \)
(c) 2
(d) 1
Answer: (b)
Sol. We have, \( 3 \sin \theta = 2 \)
\( \Rightarrow \sin \theta = \frac{2}{3} \)
\( \Rightarrow \frac{1}{\text{cosec } \theta} = \frac{2}{3} \)
\( [\because \sin \theta = \frac{1}{\text{cosec } \theta}] \)
\( \Rightarrow \text{cosec } \theta = \frac{3}{2} \).
So, the correct option is (b).

Question. If \( x \sin (90^\circ – \theta) \cot (90^\circ – \theta) = \cos (90^\circ – \theta) \), then the value of \( x \) is :
(a) 0
(b) – 1
(c) 2
(d) 1
Answer: (d)
Sol. We have, \( x \sin (90^\circ – \theta) \cot (90^\circ – \theta) = \cos (90^\circ – \theta) \)
\( \Rightarrow x \cos \theta \tan \theta = \sin \theta \)
\( \Rightarrow x \cdot \cos \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \)
\( \Rightarrow x = 1 \).
So, the correct option is (d).

Fill in the Blanks

Question. The value of \( \sin^2 25^\circ + \sin^2 65^\circ – \tan^2 45^\circ \) is ______
Answer: 0

Question. The value of \( \theta, 0^\circ < \theta < 90^\circ \), for which \( \sin \theta = \cos \theta \) is _________
Answer: \( 45^\circ \)

True/False

Question. The value of \( 3 \text{cosec}^2 \theta – 3 \cot^2 \theta \) is 3.
Answer: True

Question. If A is an acute angle, then \( \sin A \cdot \cos A \le 1 \).
Answer: False
Explanation: If A is acute, then \( \sin A \cdot \cos A \) is always less than 1.

Very Short Answer Type Questions

Question. Express \( \sin 85^\circ + \cos 76^\circ \) in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 45^\circ \).
Answer: \( \sin 85^\circ + \cos 76^\circ = \sin (90^\circ – 5^\circ) + \cos (90^\circ – 14^\circ) = \cos 5^\circ + \sin 14^\circ \). Ans.

Question. Find the value of \( (\sin^2 33^\circ + \sin^2 57^\circ) \) 
Answer: \( \sin^2 33^\circ + \sin^2 57^\circ = \sin^2 33^\circ + \sin^2 (90^\circ – 33^\circ) = \sin^2 33^\circ + \cos^2 33^\circ = 1 \). Ans.

Question. In \( \Delta ABC \) right angled at \( C \), find the value of \( \sin (A + B) \).
Answer: In \( \Delta ABC \),
\( \angle A + \angle B + \angle C = 180^\circ \)
\( \Rightarrow \angle A + \angle B + 90^\circ = 180^\circ \)
\( \Rightarrow \angle A + \angle B = 90^\circ \)
Taking \( \sin \) on both sides, we get \( \sin (A + B) = \sin 90^\circ \)
\( \Rightarrow \sin (A + B) = 1 \). Ans.

Question. If \( A + B = 90^\circ \) and \( \tan A = \frac{3}{4} \), what is the value of \( \cot B \)?
Answer: Given : \( A + B = 90^\circ \)
\( \Rightarrow B = 90^\circ – A \)
or \( \cot B = \cot (90^\circ – A) \)
\( \Rightarrow \cot B = \tan A = \frac{3}{4} \). Ans.

Question. What happens to value of \( \cos \theta \) when \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \)?
Answer: \( \cos 0^\circ = 1 \) and \( \cos 90^\circ = 0 \). So, when \( \theta \) increases from \( 0^\circ \) to \( 90^\circ \), the value of \( \cos \theta \) decreases from 1 to 0. Ans.

Question. If \( \sin^2 A = \frac{1}{2} \tan^2 45^\circ \), where A is an acute angle, find the value of A.
Answer: Given : \( \sin^2 A = \frac{1}{2} \tan^2 45^\circ \)
\( \Rightarrow \sin^2 A = \frac{1}{2}(1)^2 = \frac{1}{2} \)
\( \Rightarrow \sin A = \frac{1}{\sqrt{2}} = \sin 45^\circ \)
\( \Rightarrow A = 45^\circ \). Ans.

Question. Evaluate : \( \sin^2 60^\circ + 2 \tan 45^\circ – \cos^2 30^\circ \)
Answer: We know,
\( \sin 60^\circ = \frac{\sqrt{3}}{2}, \tan 45^\circ = 1 \) and \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \therefore \sin^2 60^\circ + 2 \tan 45^\circ – \cos^2 30^\circ \)
\( = \left( \frac{\sqrt{3}}{2} \right)^2 + 2(1) – \left( \frac{\sqrt{3}}{2} \right)^2 \)
\( = \frac{3}{4} + 2 – \frac{3}{4} \)
\( = 2. \) Ans.

Question. If \( x = a \cos \theta, y = b \sin \theta \), then find the value of \( b^2 x^2 + a^2 y^2 – a^2 b^2 \).
Answer: Given : \( x = a \cos \theta \) and \( y = b \sin \theta \)
Now,
\( b^2 x^2 + a^2 y^2 – a^2 b^2 = b^2(a \cos \theta)^2 + a^2(b \sin \theta)^2 – a^2 b^2 \)
\( = a^2 b^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta – a^2 b^2 \)
\( = a^2 b^2 (\cos^2 \theta + \sin^2 \theta) – a^2 b^2 \)
\( = a^2 b^2 (1) – a^2 b^2 \)
\( [\because \sin^2 \theta + \cos^2 \theta = 1] \)
\( = 0. \) Ans.

Question. Prove that : \( \sin^6 \theta + \cos^6 \theta = 1 – 3 \sin^2 \theta \cos^2 \theta \). 
Answer: Consider,
\( L.H.S. = \sin^6 \theta + \cos^6 \theta \)
\( = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 \)
\( = (\sin^2 \theta + \cos^2 \theta)^3 – 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) \)
\( = 1^3 – 3 \sin^2 \theta \cos^2 \theta (1) \)
\( = 1 – 3 \sin^2 \theta \cos^2 \theta = R.H.S. \) Hence Proved.

Question. Evaluate : \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \).
Answer: \( \sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ \)
\( = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} \)
\( = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \) Ans.

Question. Evaluate : \( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \) [NCERT]
Answer: \( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \)
\( = \frac{5 \left( \frac{1}{2} \right)^2 + 4 \left( \frac{2}{\sqrt{3}} \right)^2 – (1)^2}{1} \)
\( [\because \sin^2 \theta + \cos^2 \theta = 1] \)
\( = \frac{5}{4} + \frac{16}{3} – 1 \)
\( = \frac{15 + 64 – 12}{12} = \frac{67}{12}. \) Ans.

Question. If \( \sec 4A = \text{cosec } (A – 20^\circ) \), where \( 4A \) is an acute angle, find the value of \( A \).
Answer: Given : \( \sec 4A = \text{cosec } (A – 20^\circ) \)
\( \Rightarrow \text{cosec } (90^\circ – 4A) = \text{cosec } (A – 20^\circ) \)
\( [\because \sec \theta = \text{cosec } (90^\circ – \theta)] \)
\( \Rightarrow 90^\circ – 4A = A – 20^\circ \)
\( \Rightarrow 110^\circ = 5A \)
\( \Rightarrow A = \frac{110^\circ}{5} = 22^\circ \) Ans.

Question. Show that \( \tan^4 \theta + \tan^2 \theta = \sec^4 \theta – \sec^2 \theta \) 
Answer: Consider,
\( L.H.S. = \tan^4 \theta + \tan^2 \theta \)
\( = \tan^2 \theta (\tan^2 \theta + 1) \)
\( = (\sec^2 \theta – 1) \sec^2 \theta \)
\( [\because \sec^2 \theta – \tan^2 \theta = 1] \)
\( = \sec^4 \theta – \sec^2 \theta = R.H.S. \)
Hence Proved.

Question. Evaluate : \( (\sec A + \tan A) (1 – \sin A) \).
Answer: \( (\sec A + \tan A) (1 – \sin A) \)
\( = \sec A + \tan A – \sin A \cdot \sec A – \sin A \cdot \tan A \)
\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} – \sin A \cdot \frac{1}{\cos A} – \sin A \cdot \frac{\sin A}{\cos A} \)
\( = \frac{1 + \sin A – \sin A – \sin^2 A}{\cos A} \)
\( = \frac{1 – \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A \) Ans.

Question. If \( \sin \theta + \sin^2 \theta = 1 \), prove that : \( \cos^2 \theta + \cos^4 \theta = 1 \).  
Answer: Given,
\( \sin \theta + \sin^2 \theta = 1 \)
\( \Rightarrow \sin \theta + (1 – \cos^2 \theta) = 1 \)
\( \Rightarrow \sin \theta = \cos^2 \theta \)
\( \Rightarrow \sin^2 \theta = \cos^4 \theta \)
\( \Rightarrow 1 – \cos^2 \theta = \cos^4 \theta \)
\( \Rightarrow \cos^2 \theta + \cos^4 \theta = 1. \) Hence Proved.

Short Answer Type Questions-I

Question. Solve : \( \frac{\sin 50^\circ}{\cos 40^\circ} + \frac{\text{cosec } 40^\circ}{\sec 50^\circ} – 4 \cos 50^\circ \text{cosec } 40^\circ \).
Answer: \( \frac{\sin 50^\circ}{\cos 40^\circ} + \frac{\text{cosec } 40^\circ}{\sec 50^\circ} – 4 \cos 50^\circ \text{cosec } 40^\circ \)
\( = \frac{\sin (90^\circ – 40^\circ)}{\cos 40^\circ} + \frac{\text{cosec } (90^\circ – 50^\circ)}{\sec 50^\circ} – 4 \cos 50^\circ \text{cosec } (90^\circ – 50^\circ) \)
\( = \frac{\cos 40^\circ}{\cos 40^\circ} + \frac{\sec 50^\circ}{\sec 50^\circ} – 4 \cos 50^\circ \sec 50^\circ \)
\( = 1 + 1 – 4 = – 2. \) Ans.

Question. Solve : \( \frac{2}{3}(\cos^4 30^\circ – \sin^4 45^\circ) – 3(\sin^2 60^\circ – \sec^2 45^\circ) + \frac{1}{4}(\cot^2 30^\circ) \).
Answer: \( \frac{2}{3}(\cos^4 30^\circ – \sin^4 45^\circ) – 3(\sin^2 60^\circ – \sec^2 45^\circ) + \frac{1}{4}(\cot^2 30^\circ) \)
\( = \frac{2}{3} \left[ \left( \frac{\sqrt{3}}{2} \right)^4 – \left( \frac{1}{\sqrt{2}} \right)^4 \right] – 3 \left[ \left( \frac{\sqrt{3}}{2} \right)^2 – (\sqrt{2})^2 \right] + \frac{1}{4}(\sqrt{3})^2 \)
\( = \frac{2}{3} \left[ \frac{9}{16} – \frac{1}{4} \right] – 3 \left[ \frac{3}{4} – 2 \right] + \frac{3}{4} \)
\( = \frac{5}{24} + \frac{15}{4} + \frac{3}{4} = \frac{5}{24} + \frac{9}{2} \)
\( = \frac{5 + 108}{24} = \frac{113}{24}. \) Ans.

Question. Solve : \( \frac{\sec^2 \theta – \cot^2 (90^\circ – \theta)}{\text{cosec}^2 67^\circ – \tan^2 23^\circ} + \sin^2 40^\circ + \sin^2 50^\circ \).
Answer: \( \frac{\sec^2 \theta – \cot^2 (90^\circ – \theta)}{\text{cosec}^2 67^\circ – \tan^2 23^\circ} + \sin^2 40^\circ + \sin^2 50^\circ \)
\( = \frac{\sec^2 \theta – \tan^2 \theta}{\text{cosec}^2 67^\circ – \tan^2 (90^\circ – 67^\circ)} + \sin^2 40^\circ + \sin^2 (90^\circ – 40^\circ) \)
\( = \frac{\sec^2 \theta – \tan^2 \theta}{\text{cosec}^2 67^\circ – \cot^2 67^\circ} + \sin^2 40^\circ + \cos^2 40^\circ \)
\( = 1 + 1 = 2. \) Ans.

Question. Evaluate : \( \frac{2 \sin^2 63^\circ + 1 + 2 \sin^2 27^\circ}{3 \cos^2 17^\circ – 2 + 3 \cos^2 73^\circ} \).
Answer: \( \frac{2 \sin^2 63^\circ + 1 + 2 \sin^2 27^\circ}{3 \cos^2 17^\circ – 2 + 3 \cos^2 73^\circ} \)
\( = \frac{2 \sin^2 63^\circ + 1 + 2 \sin^2 (90^\circ – 63^\circ)}{3 \cos^2 17^\circ – 2 + 3 \cos^2 (90^\circ – 17^\circ)} \)
\( = \frac{2 \sin^2 63^\circ + 1 + 2 \cos^2 63^\circ}{3 \cos^2 17^\circ – 2 + 3 \sin^2 17^\circ} \)
\( = \frac{2 + 1}{3 – 2} = 3. \) Ans.

Question. Evaluate : \( \frac{\cos 58^\circ}{\sin 32^\circ} + \frac{2}{\sqrt{3}} \tan 17^\circ \tan 38^\circ \tan 60^\circ \tan 52^\circ \tan 73^\circ – 3(\sin^2 31^\circ + \sin^2 59^\circ) \).
Answer: \( \frac{\cos 58^\circ}{\sin 32^\circ} + \frac{2}{\sqrt{3}} \tan 17^\circ \tan 38^\circ \tan 60^\circ \tan 52^\circ \tan 73^\circ – 3(\sin^2 31^\circ + \sin^2 59^\circ) \)
\( = \frac{\cos (90^\circ – 32^\circ)}{\sin 32^\circ} + \frac{2}{\sqrt{3}} \tan (90^\circ – 73^\circ) \tan (90^\circ – 52^\circ) \tan 60^\circ \tan 52^\circ \tan 73^\circ – 3[\sin^2 (90^\circ – 59^\circ) + \sin^2 59^\circ] \)
\( = \frac{\sin 32^\circ}{\sin 32^\circ} + \frac{2}{\sqrt{3}} \cot 73^\circ \cot 52^\circ \tan 60^\circ \tan 52^\circ \tan 73^\circ – 3[\cos^2 59^\circ + \sin^2 59^\circ] \)
\( = 1 + \frac{2}{\sqrt{3}} \times \sqrt{3} – 3 = 1 + 2 – 3 = 0. \) Ans.

Question. If \( \sqrt{3} \sin \theta = \cos \theta \), find the value of \( \frac{3 \cos^2 \theta + 2 \cos \theta}{3 \cos \theta + 2} \).
Answer: \( \sqrt{3} \sin \theta = \cos \theta \) [Given]
\( \Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \)
or \( \tan \theta = \frac{1}{\sqrt{3}} \)
\( \Rightarrow \tan \theta = \tan 30^\circ \)
\( \Rightarrow \theta = 30^\circ \)
Now, \( \frac{3 \cos^2 \theta + 2 \cos \theta}{3 \cos \theta + 2} = \frac{\cos \theta (3 \cos \theta + 2)}{(3 \cos \theta + 2)} = \cos \theta \)
Put \( \theta = 30^\circ \)
\( \Rightarrow \cos 30^\circ = \frac{\sqrt{3}}{2} \) Ans.

Question. If \( A, B \) and \( C \) are interior angles of \( \Delta ABC \), then prove that : \( \text{cosec } \left( \frac{A + C}{2} \right) = \sec \frac{B}{2} \). 
Answer: In \( \Delta ABC \)
\( \angle A + \angle B + \angle C = 180^\circ \) [Angle sum property]
\( \Rightarrow \angle A + \angle C = 180^\circ – \angle B \)
Divide by 2 on both sides
\( \frac{A + C}{2} = \frac{180^\circ}{2} – \frac{B}{2} \)
\( \frac{A + C}{2} = 90^\circ – \frac{B}{2} \)
Taking \( \text{cosec} \) both sides,
\( \text{cosec } \left( \frac{A + C}{2} \right) = \text{cosec } \left( 90^\circ – \frac{B}{2} \right) \)
\( \text{cosec } \left( \frac{A + C}{2} \right) = \sec \frac{B}{2} \)
\( [\because \text{cosec } (90^\circ – \theta) = \sec \theta] \)
Hence Proved.

Question. In \( \Delta ABC \), if \( \angle C = 90^\circ \), prove that \( \sin^2 A + \sin^2 B = 1 \).
Answer: In \( \Delta ABC \)
\( \angle A + \angle B + \angle C = 180^\circ \) [Angle-sum property]
\( \Rightarrow \angle A + \angle B + 90^\circ = 180^\circ \)
\( \Rightarrow \angle A + \angle B = 90^\circ \)
\( \Rightarrow \angle B = 90^\circ - \angle A \)
Now, \( L.H.S. = \sin^2 A + \sin^2 B \)
\( = \sin^2 A + \sin^2 (90^\circ - A) \)
\( = \sin^2 A + \cos^2 A \)
\( = 1 = R.H.S. \)
Hence Proved.

Question. If \( \cot \theta = \frac{15}{8} \), then evaluate \( \frac{(2 + 2\sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(2 - 2\cos \theta)} \).
Answer: Sol. Given, \( \cot \theta = \frac{15}{8} \)
Now, \( \frac{(2 + 2\sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(2 - 2\cos \theta)} \)
\( = \frac{2(1 + \sin \theta)(1 - \sin \theta)}{2(1 + \cos \theta)(1 - \cos \theta)} \)
\( = \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \)
\( = \cot^2 \theta \)
\( = \left(\frac{15}{8}\right)^2 = \frac{225}{64} \). Ans.

Question. Prove that \( \frac{\cot^2 \alpha}{1 + \text{cosec } \alpha} = \text{cosec } \alpha - 1 \).  
Answer: Sol. Consider, L.H.S. \( = \frac{\cot^2 \alpha}{1 + \text{cosec } \alpha} \)
\( = \frac{\text{cosec}^2 \alpha - 1}{1 + \text{cosec } \alpha} [\because \cot^2 \alpha = \text{cosec}^2 \alpha - 1] \)
\( = \frac{(\text{cosec } \alpha + 1)(\text{cosec } \alpha - 1)}{1 + \text{cosec } \alpha} \)
\( = \text{cosec } \alpha - 1 \)
\( = \text{cosec } \alpha = R.H.S. \) Hence Proved.

Question. Prove that : \( (\sin \theta - \text{cosec } \theta)(\cos \theta - \sec \theta) = \frac{1}{\tan \theta + \cot \theta} \).
Answer: Sol. Consider, L.H.S. \( = (\sin \theta - \text{cosec } \theta)(\cos \theta - \sec \theta) \)
\( = \left(\sin \theta - \frac{1}{\sin \theta}\right) \left(\cos \theta - \frac{1}{\cos \theta}\right) \)
\( = \frac{(\sin^2 \theta - 1)(\cos^2 \theta - 1)}{\sin \theta \cos \theta} \)
\( = \frac{(-\cos^2 \theta)(-\sin^2 \theta)}{\sin \theta \cos \theta} \)
\( = \sin \theta \cos \theta \)
R.H.S. \( = \frac{1}{\tan \theta + \cot \theta} \)
\( = \frac{1}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}} \)
\( = \frac{\sin \theta \cos \theta}{\sin^2 \theta + \cos^2 \theta} \)
\( = \sin \theta \cos \theta [\because \sin^2 \theta + \cos^2 \theta = 1] \)
Thus, L.H.S. \( = \) R.H.S. \( = \sin \theta \cos \theta \).
Hence Proved.

Question. Prove that : \( (\sin \theta + \text{cosec } \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta \). 
Answer: Sol. Consider, L.H.S. \( = (\sin \theta + \text{cosec } \theta)^2 + (\cos \theta + \sec \theta)^2 \)
\( = \sin^2 \theta + \text{cosec}^2 \theta + 2 \sin \theta \text{cosec } \theta + \cos^2 \theta + \sec^2 \theta + 2 \cos \theta \sec \theta \)
\( = (\sin^2 \theta + \cos^2 \theta) + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) + 2(\sin \theta \text{cosec } \theta + \cos \theta \sec \theta) \)
\( = 1 + 1 + \cot^2 \theta + 1 + \tan^2 \theta + 4 \)
\( = 7 + \cot^2 \theta + \tan^2 \theta \)
\( = R.H.S. \) Hence Proved.

Question. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \), find the value of \( A \) and \( B \) if \( 0^\circ < A + B < 90^\circ \) and \( A > B \).
Answer: Sol. Given, \( \tan(A + B) = \sqrt{3} = \tan 60^\circ \)
or \( A + B = 60^\circ \) ...(i)
and \( \tan(A - B) = \frac{1}{\sqrt{3}} = \tan 30^\circ \)
or \( A - B = 30^\circ \) ...(ii)
Adding (i) and (ii),
\( 2A = 90^\circ \)
or \( A = 45^\circ \)
From (i), \( A + B = 60^\circ \)
or \( B = 60^\circ - 45^\circ = 15^\circ \)
Hence \( A = 45^\circ \) and \( B = 15^\circ \). Ans.

Question. Prove that : \( \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} + \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} = 2 \sec \theta \).
Answer: Sol. Consider, L.H.S. \( = \sqrt{\frac{1 + \sin \theta}{1 - \sin \theta}} + \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \)
\( = \sqrt{\frac{(1 + \sin \theta)(1 + \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}} + \sqrt{\frac{(1 - \sin \theta)(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}} \)
\( = \sqrt{\frac{(1 + \sin \theta)^2}{1 - \sin^2 \theta}} + \sqrt{\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}} \)
\( = \sqrt{\frac{(1 + \sin \theta)^2}{\cos^2 \theta}} + \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}} \)
\( = \frac{1 + \sin \theta}{\cos \theta} + \frac{1 - \sin \theta}{\cos \theta} \)
\( = \frac{1 + \sin \theta + 1 - \sin \theta}{\cos \theta} \)
\( = \frac{2}{\cos \theta} = 2 \sec \theta = R.H.S. \) Hence Proved.

Question. If \( x = 3 \sin \theta \) and \( y = 4 \cos \theta \), find the value of \( \sqrt{16x^2 + 9y^2} \).
Answer: Sol. Given, \( x = 3 \sin \theta \)
\( \Rightarrow x^2 = 9 \sin^2 \theta \)
\( \Rightarrow \sin^2 \theta = \frac{x^2}{9} \) ...(i)
and \( y = 4 \cos \theta \)
\( \Rightarrow y^2 = 16 \cos^2 \theta \)
\( \Rightarrow \cos^2 \theta = \frac{y^2}{16} \) ...(ii)
On adding equations (i) and (ii)
\( \sin^2 \theta + \cos^2 \theta = \frac{x^2}{9} + \frac{y^2}{16} \)
\( 1 = \frac{16x^2 + 9y^2}{144} [\because \sin^2 \theta + \cos^2 \theta = 1] \)
\( 16x^2 + 9y^2 = 144 \)
Taking square root of both sides
\( \sqrt{16x^2 + 9y^2} = \sqrt{144} \)
Thus, \( \sqrt{16x^2 + 9y^2} = 12 \). Ans.

Question. If \( \sec \theta + \tan \theta = m \), show that : \( \frac{m^2 - 1}{m^2 + 1} = \sin \theta \).
Answer: Sol. Given, \( \sec \theta + \tan \theta = m \)
Thus, \( \frac{m^2 - 1}{m^2 + 1} = \frac{(\sec \theta + \tan \theta)^2 - 1}{(\sec \theta + \tan \theta)^2 + 1} \)
\( = \frac{\sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta - 1}{\sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta + 1} \)
\( = \frac{(\sec^2 \theta - 1) + \tan^2 \theta + 2 \sec \theta \tan \theta}{\sec^2 \theta + (\tan^2 \theta + 1) + 2 \sec \theta \tan \theta} \)
\( = \frac{\tan^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta}{\sec^2 \theta + \sec^2 \theta + 2 \sec \theta \tan \theta} \)
\( = \frac{2 \tan^2 \theta + 2 \sec \theta \tan \theta}{2 \sec^2 \theta + 2 \sec \theta \tan \theta} \)
\( = \frac{2 \tan \theta (\tan \theta + \sec \theta)}{2 \sec \theta (\sec \theta + \tan \theta)} \)
\( = \frac{\tan \theta}{\sec \theta} = \frac{\sin \theta / \cos \theta}{1 / \cos \theta} \)
\( = \sin \theta \). Hence Proved.

Question. If \( \sin \theta + \cos \theta = \sqrt{2} \), prove that \( \tan \theta + \cot \theta = 2 \). 
Answer: Sol. Given : \( \sin \theta + \cos \theta = \sqrt{2} \)
Squaring both sides,
\( (\sin \theta + \cos \theta)^2 = (\sqrt{2})^2 \)
\( \Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2 \)
\( 1 + 2 \sin \theta \cos \theta = 2 [\because \sin^2 \theta + \cos^2 \theta = 1] \)
\( \Rightarrow 2 \sin \theta \cos \theta = 1 \)
\( \Rightarrow \sin \theta \cos \theta = \frac{1}{2} \) ...(i)
We have to prove that \( \tan \theta + \cot \theta = 2 \)
Taking L.H.S.,
\( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \)
\( = \frac{1}{\sin \theta \cos \theta} = \frac{1}{1/2} = 2 = R.H.S. \) Hence Proved.

Question. If \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \), show that \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \).
Answer: Sol. Given, \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \)
\( \Rightarrow (\cos \theta + \sin \theta)^2 = 2 \cos^2 \theta \)
\( \Rightarrow \cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = 2 \cos^2 \theta \)
\( \Rightarrow 2 \cos^2 \theta - \cos^2 \theta - \sin^2 \theta = 2 \sin \theta \cos \theta \)
\( \Rightarrow \cos^2 \theta - \sin^2 \theta = 2 \sin \theta \cos \theta \)
\( \Rightarrow (\cos \theta + \sin \theta)(\cos \theta - \sin \theta) = 2 \sin \theta \cos \theta \)
\( \Rightarrow \sqrt{2} \cos \theta (\cos \theta - \sin \theta) = 2 \sin \theta \cos \theta \)
\( \Rightarrow \cos \theta - \sin \theta = \sqrt{2} \sin \theta \).
Hence Proved.

Question. Prove that : \( (1 - \sin \theta + \cos \theta)^2 = 2(1 + \cos \theta)(1 - \sin \theta) \).
Answer: Sol. Consider,
L.H.S. \( = (1 - \sin \theta + \cos \theta)^2 \)
\( = [1 - (\sin \theta - \cos \theta)]^2 \)
\( = 1 + (\sin \theta - \cos \theta)^2 - 2(1)(\sin \theta - \cos \theta) \)
\( = 1 + \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta - 2 \sin \theta + 2 \cos \theta \)
\( = 1 + 1 - 2 \sin \theta \cos \theta - 2 \sin \theta + 2 \cos \theta \)
\( = 2 - 2 \sin \theta \cos \theta - 2 \sin \theta + 2 \cos \theta \)
\( = 2[1 - \sin \theta \cos \theta - \sin \theta + \cos \theta] \)
R.H.S. \( = 2(1 + \cos \theta)(1 - \sin \theta) \)
\( = 2[1 - \sin \theta + \cos \theta - \sin \theta \cos \theta] \)
\( = 2[1 - \sin \theta \cos \theta - \sin \theta + \cos \theta] \)
L.H.S. \( = \) R.H.S. Hence Proved.

Question. If tan 2A = cot (A – 18°), wehre 2A is an acute angle, find the value of A. 
Answer: Given: \( \tan 2A = \cot (A - 18^\circ) \), \( 0 \le 2A < 90^\circ \). (2A is acute)
To find: value of A.
We know, \( \tan \theta = \cot (90^\circ - \theta) \) and \( \cot \theta = \tan (90^\circ - \theta) \).
\( \rightarrow \cot (90^\circ - 2A) = \cot (A - 18^\circ) \)
Applying \( \cot^{-1} \) on both sides,
\( 90^\circ - 2A = A - 18^\circ \)
\( 108^\circ = 3A \)
\( \rightarrow A = 36^\circ \).
The value of A is \( 36^\circ \).

Question. If \( \sin \theta + \cos \theta = \sqrt{3} \), then prove that \( \tan \theta + \cot \theta = 1 \). 
Answer: Given : \( \sin \theta + \cos \theta = \sqrt{3} \)
Squaring both sides,
\( (\sin \theta + \cos \theta)^2 = (\sqrt{3})^2 \)
\( \Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3 \)
\( \Rightarrow 1 + 2 \sin \theta \cos \theta = 3 \) [\( \because \sin^2 \theta + \cos^2 \theta = 1 \)]
\( \Rightarrow 2 \sin \theta \cos \theta = 2 \)
\( \Rightarrow \sin \theta \cos \theta = 1 \)
Now, \( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \)
\( = \frac{1}{1} = 1 \). Hence Proved.

Question. If \( \sin (A + 2B) = \frac{\sqrt{3}}{2} \) and \( \cos (A + 4B) = 0, A > B \) and \( A + 4B \le 90^\circ \) then find A and B. [20
Answer: Given : \( \sin (A + 2B) = \frac{\sqrt{3}}{2} \) and \( \cos (A + 4B) = 0 \)
\( \Rightarrow \sin (A + 2B) = \sin 60^\circ \) and \( \cos (A + 4B) = \cos 90^\circ \)
\( \Rightarrow A + 2B = 60^\circ \) and \( A + 4B = 90^\circ \)
On subtracting the above equations, we get
\( 2B = 30^\circ \)
\( \Rightarrow B = \frac{30^\circ}{2} = 15^\circ \)
So, \( A = 60^\circ – 2B \)
\( = 60^\circ – 2 \times 15^\circ \)
\( = 30^\circ \)
\( \therefore A = 30^\circ \) and \( B = 15^\circ \).

Question. Without using trigonometric tables, evaluate : \( \frac{\cos 58^\circ}{\sin 32^\circ} + \frac{\sin 22^\circ}{\cos 68^\circ} - \frac{\cos 38^\circ \text{cosec } 52^\circ}{\tan 18^\circ \tan 35^\circ \tan 60^\circ \tan 72^\circ \tan 55^\circ} \)
Answer: We have,
\( \frac{\cos 58^\circ}{\sin 32^\circ} + \frac{\sin 22^\circ}{\cos 68^\circ} - \frac{\cos 38^\circ \text{cosec } 52^\circ}{\tan 18^\circ \tan 35^\circ \tan 60^\circ \tan 72^\circ \tan 55^\circ} \)
\( = \frac{\cos (90^\circ - 32^\circ)}{\sin 32^\circ} + \frac{\sin (90^\circ - 68^\circ)}{\cos 68^\circ} - \frac{\cos (90^\circ - 52^\circ) \text{cosec } 52^\circ}{\tan (90^\circ - 72^\circ) \tan (90^\circ - 55^\circ) \tan 60^\circ \tan 72^\circ \tan 55^\circ} \)
\( = \frac{\sin 32^\circ}{\sin 32^\circ} + \frac{\cos 68^\circ}{\cos 68^\circ} - \frac{\sin 52^\circ \text{cosec } 52^\circ}{\cot 72^\circ \cot 55^\circ \tan 60^\circ \tan 72^\circ \tan 55^\circ} \)
[\( \because \cos (90^\circ - \theta) = \sin \theta \), \( \sin (90^\circ - \theta) = \cos \theta \), \( \tan (90^\circ - \theta) = \cot \theta \)]
\( = 1 + 1 - \frac{1}{\sqrt{3}} = 2 - \frac{1}{\sqrt{3}} \).

Question. Solve : \( \frac{3 \cos 55^\circ}{7 \sin 35^\circ} - \frac{4(\cos 70^\circ \text{cosec } 20^\circ)}{7(\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ)} \)
Answer: We have,
\( \frac{3 \cos 55^\circ}{7 \sin 35^\circ} - \frac{4(\cos 70^\circ \text{cosec } 20^\circ)}{7(\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ)} \)
\( = \frac{3 \cos (90^\circ - 35^\circ)}{7 \sin 35^\circ} - \frac{4 \cos (90^\circ - 20^\circ) \text{cosec } 20^\circ}{7 \tan (90^\circ - 85^\circ) \tan (90^\circ - 65^\circ) \tan 45^\circ \tan 65^\circ \tan 85^\circ} \)
\( = \frac{3 \sin 35^\circ}{7 \sin 35^\circ} - \frac{4 \sin 20^\circ \text{cosec } 20^\circ}{7 \cot 85^\circ \cot 65^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \)
\( = \frac{3}{7} - \frac{4}{7 \tan 45^\circ} = \frac{3}{7} - \frac{4}{7} = - \frac{1}{7} \).