## Prove the following identities,

[question] Question. Prove the following identities, where the angles involved are acute angles for which the following expressions are defined. (i) $(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$. (ii) $\frac{\cos \mathbf{A}}{\mathbf{1}+\sin \mathbf{A}}+\frac{\mathbf{1}+\sin \mathbf{A}}{\cos \mathbf{A}}=2 \sec \mathrm{A}$. (iii) $\frac{\tan \theta}{\mathbf{1}-\cot \theta}+\frac{\cot \theta}{\mathbf{1}-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$ (i...

## Choose the correct option. Justify your choice :

[question] Question. Choose the correct option. Justify your choice : (i) $9 \sec ^{2} A-9 \tan ^{2} A=$ (A) 1 (B) 9 (C) 8 (D) 0 (ii) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$ (A) 0 (B) 1 (C) 2 (D) – 1 (iii) $(\sec A+\tan A)(1-\sin A)=$ (A) sec A (B) sin A (C) cosec A (D) cos A (iv) $\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=$ (A) $\sec ^{2} \mathrm{~A}$ (B) $-1$ (C) $\cot ^{2} A$ (D) $\tan ^{2} \mathrm{~A}$ [/question] [solution] Sol...

## Write all the other trigonometric ratios of $\angle \mathrm{A}$ in terms of $\sec \mathrm{A}$.

[question] Question. Write all the other trigonometric ratios of $\angle \mathrm{A}$ in terms of $\sec \mathrm{A}$. [/question] [solution] Solution: (i) $\sin A=\sqrt{1-\cos ^{2} A}$ $=\sqrt{1-\frac{1}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A}$ (ii) $\cos A=\frac{1}{\sec A}$ (iii) $\tan A=\sqrt{\sec ^{2} \mathbf{A}-\mathbf{1}}$ (iv) $\cot A=\frac{1}{\tan A}=\frac{1}{\sqrt{\sec ^{2} A-1}}$ (v) $\operatorname{cosec} A=\frac{1}{\sin A}=\frac{\sec A}{\sqrt{\sec ^{2} A-1}}$ [/solution]...

## Express the trigonometric ratios sin A,

[question] Question. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. [/question] [solution] Solution: We have $\operatorname{cosec}^{2} A-\cot ^{2} A=1$ $\Rightarrow \operatorname{cosec}^{2} A=1+\cot ^{2} A$ $\Rightarrow(\operatorname{cosec} A)^{2}=\cot ^{2} A+1$ $\Rightarrow\left(\frac{1}{\sin A}\right)^{2}=\cot ^{2} A+1$ $\Rightarrow(\sin \mathrm{A})^{2}=\frac{\mathbf{1}}{\cot ^{2} \mathbf{A}+\mathbf{1}}$ $\Rightarrow \sin A=\pm \frac{1}{\sqrt{\cot ^{2} A+1}}$ We rej...

## Express $\sin 67^{\circ}+\cos 75^{\circ}$

[question] Question. Express $\sin 67^{\circ}+\cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$. [/question] [solution] Solution: sin 67° + cos 75° = sin(90° – 23°) + cos (90° – 15°) = cos 23° + sin 15° [/solution]...

## If tan A = cot B, prove that A + B = 90°.

[question] Question. If tan A = cot B, prove that A + B = 90°. [/question] [solution] Solution: tan A = cot B tan A = tan (90°–B) $\therefore \mathrm{A}=90^{\circ}-\mathrm{B}$ A + B = 90° [/solution]...

## If tan 2 A = cot (A – 18°), where 2 A is an acute angle,

[question] Question. If tan 2 A = cot (A – 18°), where 2 A is an acute angle, find the value of A. [/question] [solution] Solution: tan2A = cot(A – 18°) cot (90° – 2A) = cot (A – 18°) 90° – 2A = A – 18° 108° = 3A A = 36° [/solution]...

## State whether the following are true or false. Justify your answer

[question] Question. State whether the following 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) $\cot \mathrm{A}$ is not defined for $\mathrm{A}=0^{\circ}$. [/question] [solution] Solution: (i) False. When A = 60°, B = 30° LHS = sin (A + B) = sin (60° + 30°) = sin 90° = 1 RHS = si...

## State whether the following are true or false.

[question] Question. State whether the following are true or false. Justify your answer (i) The value of tan A is always less than 1. (ii) $\sec A=\frac{12}{5}$ for some value of angle $A$. (iii) cos A is the abbreviation used for the cosecant of angle A. (iv) cot A is the product of cot and A. (v) $\sin \theta=\frac{4}{3}$ for some angle $\theta$. [/question] [solution] Solution: (i) False. We know that $60^{\circ}=\sqrt{\mathbf{3}}>\mathbf{1}$. (ii) True. We know that value of $\sec A$ is alwa...