CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 09

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

Question. Evaluate the following: \( 2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ \)
Answer: \( 2 \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ \)
\( = 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( = 2 + \frac{3}{4} - \frac{3}{4} = 2 \)

Question. Evaluate the following: \( \frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ} \)
Answer: \( \frac{\cos 45^\circ}{\sec 30^\circ + \text{cosec } 30^\circ} \)
\( = \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + \frac{2}{1}} = \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} \)
\( = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})} \)
Multiplying numerator and denominator by \( (1 - \sqrt{3}) \):
\( = \frac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2}(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2}(1 - 3)} \)
\( = \frac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2}(-2)} = \frac{\sqrt{3}(1 - \sqrt{3})}{-4\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( = \frac{-\sqrt{6}(1 - \sqrt{3})}{8} = \frac{-\sqrt{6} + \sqrt{18}}{8} = \frac{-\sqrt{6} + 3\sqrt{2}}{8} = \frac{3\sqrt{2} - \sqrt{6}}{8} \)

Question. Evaluate the following: \( \frac{\sin 30^\circ + \tan 45^\circ - \text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ} \)
Answer: \( \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1} = \frac{\frac{3}{2} - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{3}{2}} = \frac{\frac{3\sqrt{3} - 4}{2\sqrt{3}}}{\frac{4 + 3\sqrt{3}}{2\sqrt{3}}} \)
\( = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4} \)
Multiplying numerator and denominator by \( (3\sqrt{3} - 4) \):
\( = \frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)} = \frac{27 + 16 - 24\sqrt{3}}{27 - 16} \)
\( = \frac{43 - 24\sqrt{3}}{11} \)

Question. Evaluate the following: \( \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \) 
Answer: \( \frac{5 \left(\frac{1}{2}\right)^2 + 4 \left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \)
\( = \frac{5 \left(\frac{1}{4}\right) + 4 \left(\frac{4}{3}\right) - 1}{\frac{1}{4} + \frac{3}{4}} = \frac{\frac{5}{4} + \frac{16}{3} - 1}{1} \)
\( = \frac{15 + 64 - 12}{12} = \frac{67}{12} \)

Question. Choose the correct option and justify your choice: \( \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} = \)
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°
Answer: (a) sin 60°
Justification: \( \frac{2 \left(\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} = \sin 60^\circ \)

Question. Choose the correct option and justify your choice: \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \)
(a) tan 90°
(b) 1
(c) sin 45°
(d) 0
Answer: (d) 0
Justification: \( \frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0 \)

Question. Choose the correct option and justify your choice: \( \sin 2A = 2 \sin A \) is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°
Answer: (a) 0°
Justification: For A = 0°, LHS \( = \sin(2 \times 0^\circ) = \sin 0^\circ = 0 \). RHS \( = 2 \sin 0^\circ = 2 \times 0 = 0 \).

Question. Choose the correct option and justify your choice: \( \frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ} = \)
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°
Answer: (c) tan 60°
Justification: \( \frac{2 \left(\frac{1}{\sqrt{3}}\right)}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} = \tan 60^\circ \)

Question. If \( \tan (A + B) = \sqrt{3} \) and \( \tan (A - B) = \frac{1}{\sqrt{3}} \); \( 0^\circ < A + B \le 90^\circ \); \( A > B \), find A and B. 
Answer: \( \tan (A + B) = \sqrt{3} \) \( \implies \) \( \tan (A + B) = \tan 60^\circ \) \( \implies \) \( A + B = 60^\circ \) …(i)
\( \tan (A - B) = \frac{1}{\sqrt{3}} \) \( \implies \) \( \tan (A - B) = \tan 30^\circ \) \( \implies \) \( A - B = 30^\circ \) …(ii)
Adding (i) and (ii), we get
\( 2A = 90^\circ \)

\( \implies \) \( A = \frac{90^\circ}{2} = 45^\circ \)
From (i), \( 45^\circ + B = 60^\circ \)

\( \implies \) \( B = 60^\circ - 45^\circ = 15^\circ \)
Hence, \( \angle A = 45^\circ, \angle B = 15^\circ \)

Question. State whether the following statements are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin \( \theta \) increases as \( \theta \) increases.
(iii) The value of cos \( \theta \) increases as \( \theta \) increases.
(iv) sin \( \theta \) = cos \( \theta \) for all values of \( \theta \).
(v) cot A is not defined for A = 0°.
Answer: (i) False. Let A = 60° and B = 30°. LHS = sin(60° + 30°) = sin 90° = 1. RHS = sin 60° + sin 30° = \( \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \). LHS \( \neq \) RHS.
(ii) True. sin 0° = 0, sin 30° = 1/2, sin 45° = \( 1/\sqrt{2} \), sin 60° = \( \sqrt{3}/2 \), sin 90° = 1. Value increases from 0 to 1.
(iii) False. cos 0° = 1, cos 30° = \( \sqrt{3}/2 \), cos 45° = \( 1/\sqrt{2} \), cos 60° = 1/2, cos 90° = 0. Value decreases from 1 to 0.
(iv) False. For \( \theta = 30^\circ \), sin 30° = 1/2 and cos 30° = \( \sqrt{3}/2 \). sin 30° \( \neq \) cos 30°.
(v) True. \( \cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0} \), which is not defined.

Question. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer: From trigonometric identity, \( \text{cosec}^2 A - \cot^2 A = 1 \), we get
\( \text{cosec}^2 A = 1 + \cot^2 A \)

\( \implies \) \( \text{cosec } A = \sqrt{1 + \cot^2 A} \)

\( \implies \) \( \sin A = \frac{1}{\sqrt{1 + \cot^2 A}} \)
Again from trigonometric identity \( \sec^2 A - \tan^2 A = 1 \)

\( \implies \) \( \sec^2 A = 1 + \tan^2 A = 1 + \frac{1}{\cot^2 A} = \frac{\cot^2 A + 1}{\cot^2 A} \)

\( \implies \) \( \sec A = \frac{\sqrt{\cot^2 A + 1}}{\cot A} \)
Since \( \tan A = \frac{1}{\cot A} \)

Question. Write all the other trigonometric ratios of \( \angle A \) in terms of sec A.
Answer: Since \( \sin^2 A + \cos^2 A = 1 \),
therefore \( \sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A} \)

\( \implies \) \( \sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A} \)
\( \cos A = \frac{1}{\sec A} \)
\( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{\sqrt{\sec^2 A - 1}}{\sec A}}{\frac{1}{\sec A}} = \sqrt{\sec^2 A - 1} \)
\( \text{cosec } A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}} \)
\( \cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}} \)

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

Question. Choose the correct option. Justify your choice: (1 + tan \( \theta \) + sec \( \theta \)) (1 + cot \( \theta \) - cosec \( \theta \)) =
(a) 0
(b) 1
(c) 2
(d) -1
Answer: (c) 2
Justification: \( \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) = \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \)
\( = \frac{(\cos \theta + \sin \theta)^2 - (1)^2}{\cos \theta \sin \theta} = \frac{\cos^2 \theta + \sin^2 \theta + 2 \cos \theta \sin \theta - 1}{\cos \theta \sin \theta} \)
\( = \frac{1 + 2 \cos \theta \sin \theta - 1}{\cos \theta \sin \theta} = \frac{2 \cos \theta \sin \theta}{\cos \theta \sin \theta} = 2 \)

Question. Choose the correct option. Justify your choice: (sec A + tan A) (1 - sin A) =
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Answer: (d) cos A
Justification: \( \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 - \sin A) = \frac{(1 + \sin A)(1 - \sin A)}{\cos A} = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A \)

Question. Choose the correct option. Justify your choice: \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \)
(a) sec² A
(b) -1
(c) cot² A
(d) tan² A
Answer: (d) tan² A
Justification: \( \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \)

Question. Prove the following identity: \( (\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta} \)
Answer: LHS \( = \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 = \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 \)
\( = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS} \)

Question. Prove the following identity: \( \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A \)
Answer: LHS \( = \frac{\cos^2 A + (1 + \sin A)^2}{\cos A(1 + \sin A)} = \frac{\cos^2 A + 1 + \sin^2 A + 2 \sin A}{\cos A(1 + \sin A)} \)
\( = \frac{(\cos^2 A + \sin^2 A) + 1 + 2 \sin A}{\cos A(1 + \sin A)} = \frac{1 + 1 + 2 \sin A}{\cos A(1 + \sin A)} = \frac{2 + 2 \sin A}{\cos A(1 + \sin A)} \)
\( = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} = \frac{2}{\cos A} = 2 \sec A = \text{RHS} \)

Question. Prove the following identity: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta \)
Answer: Converting into sin and cos:
LHS \( = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}} \)
\( = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)} = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)} \)
\( = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
\( = \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \)
\( = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \)
\( = \text{cosec } \theta \sec \theta + 1 = \text{RHS} \)

Question. Prove the following identity: \( \frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A} \)
Answer: LHS \( = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} = \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \cos A + 1 \)
RHS \( = \frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A \)
LHS = RHS.

Question. Prove the following identity: \( \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A \), using the identity \( \text{cosec}^2 A = 1 + \cot^2 A \).
Answer: LHS \( = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} \)
Dividing numerator and denominator by sin A:
\( = \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A} = \frac{\cot A + \text{cosec } A - 1}{\cot A - \text{cosec } A + 1} \)
\( = \frac{(\cot A + \text{cosec } A) - (\text{cosec}^2 A - \cot^2 A)}{\cot A - \text{cosec } A + 1} \)
\( = \frac{(\cot A + \text{cosec } A) - [(\text{cosec } A - \cot A)(\text{cosec } A + \cot A)]}{\cot A - \text{cosec } A + 1} \)
\( = \frac{(\cot A + \text{cosec } A) [1 - (\text{cosec } A - \cot A)]}{\cot A - \text{cosec } A + 1} = \frac{(\cot A + \text{cosec } A) (1 - \text{cosec } A + \cot A)}{\cot A - \text{cosec } A + 1} \)
\( = \text{cosec } A + \cot A = \text{RHS} \)

Question. Prove the following identity: \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \) [DoE]
Answer: LHS \( = \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}} \)
\( = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \)
\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS} \)

Question. Prove the following identity: \( \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta \)
Answer: LHS \( = \frac{\sin \theta(1 - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - 1)} \)
\( = \frac{\sin \theta(\sin^2 \theta + \cos^2 \theta - 2 \sin^2 \theta)}{\cos \theta(2 \cos^2 \theta - (\sin^2 \theta + \cos^2 \theta))} = \frac{\sin \theta(\cos^2 \theta - \sin^2 \theta)}{\cos \theta(\cos^2 \theta - \sin^2 \theta)} \)
\( = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS} \)

Question. Prove the following identity: \( (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A \)
Answer: LHS \( = (\sin^2 A + \text{cosec}^2 A + 2 \sin A \text{cosec } A) + (\cos^2 A + \sec^2 A + 2 \cos A \sec A) \)
\( = \sin^2 A + \cos^2 A + \text{cosec}^2 A + \sec^2 A + 2(1) + 2(1) \)
\( = 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4 = 7 + \tan^2 A + \cot^2 A = \text{RHS} \)

Question. Prove the following identity: \( (\text{cosec } A - \sin A) (\sec A - \cos A) = \frac{1}{\tan A + \cot A} \)
Answer: LHS \( = \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right) = \left(\frac{1 - \sin^2 A}{\sin A}\right) \left(\frac{1 - \cos^2 A}{\cos A}\right) \)
\( = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \cos A \sin A \)
RHS \( = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{\sin A \cos A}{1} = \sin A \cos A \)
LHS = RHS.

Question. Prove the following identity: \( \left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A \) [CBSE 2018(C)]
Answer: \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \) ...(i)
\( \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2 = \left(\frac{(1 - \tan A) \tan A}{-(1 - \tan A)}\right)^2 = (-\tan A)^2 = \tan^2 A \) ...(ii)
From (i) and (ii), LHS = RHS.

Question. If \( \Delta ABC \) is right angled at C, then the value of cos (A + B) is
(a) 0
(b) 1
(c) 1/2
(d) \( \sqrt{3}/2 \)
Answer: (a) 0
Justification: \( A + B + C = 180^\circ \). Since \( \angle C = 90^\circ \), \( A + B = 90^\circ \). cos 90° = 0.

Question. If sin \( \theta \) – cos \( \theta \) = 0, then the value of (\( \sin^4 \theta + \cos^4 \theta \)) is
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4
Answer: (c) 1/2
Justification: \( \sin \theta = \cos \theta \implies \tan \theta = 1 \implies \theta = 45^\circ \).
Value \( = (\sin 45^\circ)^4 + (\cos 45^\circ)^4 = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \).

Question. Show that \( \frac{1}{\sec x - \tan x} - \frac{1}{\cos x} = \frac{1}{\cos x} - \frac{1}{\sec x + \tan x} \).
Answer: LHS \( = \frac{\sec x + \tan x}{\sec^2 x - \tan^2 x} - \sec x = \sec x + \tan x - \sec x = \tan x \).
RHS \( = \sec x - \frac{\sec x - \tan x}{\sec^2 x - \tan^2 x} = \sec x - (\sec x - \tan x) = \tan x \).
LHS = RHS.

Question. If \(\sin \theta + \sin^2 \theta = 1\), prove that \(\cos^2 \theta + \cos^4 \theta = 1\).
Answer: \( \sin \theta = 1 - \sin^2 \theta = \cos^2 \theta \).
\( \implies \sin^2 \theta = \cos^4 \theta \).
Adding \(\sin \theta\) on both sides: \( \sin \theta + \sin^2 \theta = \cos^2 \theta + \cos^4 \theta \).
\( \implies 1 = \cos^2 \theta + \cos^4 \theta \).

Question. Prove that \( \frac{\cos A}{1 - \tan A} - \frac{\sin^2 A}{\cos A - \sin A} = \sin A + \cos A \).
Answer: LHS \( = \frac{\cos A}{1 - \frac{\sin A}{\cos A}} - \frac{\sin^2 A}{\cos A - \sin A} = \frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A} \)
\( = \frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A} = \cos A + \sin A = \) RHS.

Question. Prove that \( \sec A (1 - \sin A) (\sec A + \tan A) = 1 \).
Answer: LHS \( = (\sec A - \sec A \sin A) (\sec A + \tan A) = (\sec A - \tan A)(\sec A + \tan A) \)
\( = \sec^2 A - \tan^2 A = 1 = \) RHS.

Question. Determine the value of x such that \( 2 \text{cosec}^2 30^\circ + x \sin^2 60^\circ - \frac{3}{4} \tan^2 30^\circ = 10 \)
Answer: \( 2(2)^2 + x \left( \frac{\sqrt{3}}{2} \right)^2 - \frac{3}{4} \left( \frac{1}{\sqrt{3}} \right)^2 = 10 \)
\( 8 + \frac{3x}{4} - \frac{1}{4} = 10 \)
\( \frac{3x - 1}{4} = 2 \implies 3x - 1 = 8 \implies 3x = 9 \implies x = 3 \).

Question. If \(\theta\) is an acute angle and \(\tan \theta + \cot \theta = 2\), find the value of \(\tan^9 \theta + \cot^9 \theta\).
Answer: \( \tan \theta + \frac{1}{\tan \theta} = 2 \implies \tan^2 \theta - 2\tan \theta + 1 = 0 \)
\( \implies (\tan \theta - 1)^2 = 0 \implies \tan \theta = 1 \).
\( \cot \theta = 1/1 = 1 \).
Value \( = 1^9 + 1^9 = 2 \).

Question. If \( \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \), then find acute angle \(\theta\).
Answer: Dividing LHS by \(\cos \theta\):
\( \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \)
\( \implies \tan \theta = \sqrt{3} \)
\( \implies \theta = 60^\circ \)

Question. Evaluate: \( \frac{\cos 60^\circ + \sin 45^\circ - \cot 30^\circ}{\tan 60^\circ + \sec 45^\circ - \text{cosec } 30^\circ} \div \frac{\sin 30^\circ + \sin 45^\circ - \tan 60^\circ}{\cot 30^\circ + \text{cosec } 45^\circ - \sec 60^\circ} \)
Answer: The expression is of form \( \frac{1/2 + 1/\sqrt{2} - \sqrt{3}}{\sqrt{3} + \sqrt{2} - 2} \div \frac{1/2 + 1/\sqrt{2} - \sqrt{3}}{\sqrt{3} + \sqrt{2} - 2} \).
Since both sides of division are identical, the result is 1.

Question. If \( a \cos \theta + b \sin \theta = 5 \) and \( a \sin \theta - b \cos \theta = 4 \), find the value of \( a^2 + b^2 \)
Answer: Squaring and adding both equations:
\( (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 = 5^2 + 4^2 \)
\( a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = 25 + 16 \)
\( a^2 + b^2 = 41 \)

Question. If \(\sin \theta + \cos \theta = 1 (0^\circ < \theta < 90^\circ)\) then find the value of \(\sin^3 \theta + \cos^3 \theta\)
Answer: Squaring \( \sin \theta + \cos \theta = 1 \):
\( 1 + 2\sin \theta \cos \theta = 1 \implies \sin \theta \cos \theta = 0 \).
Since \( 0 < \theta < 90^\circ \), this is not strictly possible. Assuming boundaries are included, either \( \sin \theta = 1, \cos \theta = 0 \) or vice-versa.
Value \( = 1^3 + 0^3 = 1 \).

Question. Assertion (A): If \(\sin \theta = \frac{1}{2}\) and \(\theta\) is acute angle, then \((3 \cos \theta - 4 \cos^3 \theta)\) is equal to 0.
Reason (R): As \(\sin \theta = \frac{1}{2}\) and \(\theta\) is acute, so \(\theta\) must be 60°.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (c) A is true but R is false.

Question. Assertion (A): \( \cos^2 A - \sin^2 A = 1 \), \( \tan^2 A - \sec^2 A = 1 \) are trigonometric identities.
Reason (R): An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angles involved.

(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.