We know that:

$\sec^2 A  –  \tan^2 A = 1 $

$\implies \sec^2 A = 1 + \tan^2 A$

$\sec^2 A = 1 + \frac{1}{\cot^2 A} \quad \left[\because \tan A = \frac{1}{\cot A}\right]$

$\Rightarrow \sec^2 A = \frac{\cot^2 A + 1}{\cot^2 A}$

$\therefore \sec A = \frac{\sqrt{\cot^2 A + 1}}{\cot A}$

For $\tan A$:

$\therefore \tan A = \frac{1}{\cot A}$

Question 2

Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

Solution:

For $\sin A$:

We know that:

$\sin^2 A + \cos^2 A = 1$

$\Rightarrow \sin^2 A = 1 – \cos^2 A$

$\Rightarrow \sin^2 A = 1 – \frac{1}{\sec^2 A} \quad \left[\because \cos A = \frac{1}{\sec A}\right]$

$\Rightarrow \sin^2 A = \frac{\sec^2 A – 1}{\sec^2 A}$

$\therefore \sin A = \frac{\sqrt{\sec^2 A – 1}}{\sec A}$

For $\cos A$:

$\therefore \cos A = \frac{1}{\sec A}$

For $\tan A$:

We know that:

$1 + \tan^2 A = \sec^2 A$

$\Rightarrow \tan^2 A = \sec^2 A – 1$

$\therefore \tan A = \sqrt{\sec^2 A – 1}$

For $\cot A$:

We know that:

$\cot A = \frac{1}{\tan A}$

$\therefore \cot A = \frac{1}{\sqrt{\sec^2 A – 1}}$

For $cosec A$:

We know that:

$cosec A = \frac{1}{\sin A}$

$\Rightarrow cosec A = \frac{1}{\frac{\sqrt{\sec^2 A – 1}}{\sec A}}$

$\therefore cosec A = \frac{\sec A}{\sqrt{\sec^2 A – 1}}$

Question 3

Evaluate:

(i) $\frac{\sin^2 63^\circ + \sin^2 27^\circ}{\cos^2 17^\circ + \cos^2 73^\circ}$

Solution:

$\frac{(\sin 63^\circ)^2 + \sin^2 27^\circ}{(\cos 17^\circ)^2 + \cos^2 73^\circ}$

$= \frac{\{\sin(90^\circ – 27^\circ)\}^2 + \sin^2 27^\circ}{\{\cos(90^\circ – 73^\circ)\}^2 + \cos^2 73^\circ}$

$= \frac{\cos^2 27^\circ + \sin^2 27^\circ}{\sin^2 73^\circ + \cos^2 73^\circ}$  

$= \frac{1}{1} \quad [\text{Using } \sin^2 A + \cos^2 A = 1]$

$= 1$

(ii) $\sin 25^\circ \cos 65^\circ + \cos 25^\circ \sin 65^\circ$

Solution:

$ \sin 25^\circ \cos(90^\circ – 25^\circ) + \cos 25^\circ \sin(90^\circ – 25^\circ)$

$= \sin 25^\circ \sin 25^\circ + \cos 25^\circ \cos 25^\circ $

$= \sin^2 25^\circ + \cos^2 25^\circ$

$= 1 $

Question 4

Choose the correct option. Justify your choice.

(i) $9\sec^2 A – 9\tan^2 A =$

(A) 1

(B) 9

(C) 8

(D) 0

Ans:- (B) 9

Justification:

$= 9(\sec^2 A – \tan^2 A)$

$= 9 \times 1 \quad [\because \sec^2 A – \tan^2 A = 1]$

$= 9$

Therefore, the correct option is (B) 9.

(ii) $(1 + \tan \theta + \sec \theta)(1 + \cot \theta – \csc \theta) =$

(A) 0

(B) 1

(C) 2

(D) -1

Ans:- (C) 2

Justification:

$(1 + \tan \theta + \sec \theta)(1 + \cot \theta – \csc \theta) =$

$= \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)\left(1 + \frac{\cos \theta}{\sin \theta} – \frac{1}{\sin \theta}\right)$

$= \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)\left(\frac{\sin \theta + \cos \theta – 1}{\sin \theta}\right)$

$= \frac{\{(\sin \theta + \cos \theta) + 1\}\{(\sin \theta + \cos \theta) – 1\}}{\cos \theta \sin \theta}$

$= \frac{(\sin \theta + \cos \theta)^2 – (1)^2}{\cos \theta \sin \theta} $

$= \frac{\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta – 1}{\sin \theta \cos \theta} $

$= \frac{1 + 2\sin \theta \cos \theta – 1}{\sin \theta \cos \theta} $

$= \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$= 2$

Therefore, the correct option is (C) 2.

(iii) $(\sec A + \tan A)(1 – \sin A) =$

(A) $\sec A$

(B) $\sin A$

(C) $\csc A$

(D) $\cos A$

Ans:- (D) $\cos A$

Justification:

$(\sec A + \tan A)(1 – \sin A) =$

$= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 – \sin A)$

$= \left(\frac{1 + \sin A}{\cos A}\right)(1 – \sin A)$

$= \frac{(1 + \sin A)(1 – \sin A)}{\cos A}$

$= \frac{1^2 – \sin^2 A}{\cos A} $

$= \frac{1 – \sin^2 A}{\cos A}$

$= \frac{\cos^2 A}{\cos A} $

$= \cos A$

Therefore, the correct option is (D) $\cos A$.

(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$

(A) $\sec^2 A$

(B) -1

(C) $\cot^2 A$

(D) $\tan^2 A$

Ans:- (D) $\tan^2 A$

Justification:

$\frac{1 + \tan^2 A}{1 + \cot^2 A} =$

$= \frac{\sec^2 A}{cosec^2 A} $

$= \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} $

$= \frac{\sin^2 A}{\cos^2 A}$

$= \tan^2 A$

Therefore, the correct option is (D) $\tan^2 A$.

Question 5: Prove the Identities

(i) $(cosec \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta}$

Solution:

$\text{LHS} = (cosec \theta – \cot \theta)^2$

$= \left(\frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta}\right)^2$

$= \left(\frac{1 – \cos \theta}{\sin \theta}\right)^2$

$= \frac{(1 – \cos \theta)^2}{\sin^2 \theta}$

$= \frac{(1 – \cos \theta)^2}{1 – \cos^2 \theta} $

$= \frac{(1 – \cos \theta)^2}{1^2 – \cos^2 \theta}$

$= \frac{(1 – \cos \theta)(1 – \cos \theta)}{(1 + \cos \theta)(1 – \cos \theta)} $

$= \frac{1 – \cos \theta}{1 + \cos \theta}$

$= \text{RHS}$

$\therefore\text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2\sec A$

Solution:

$\text{LHS} = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$

$= \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}$

$= \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1 + \sin A)\cos A} $

$= \frac{(\cos^2 A + \sin^2 A) + 1 + 2\sin A}{(1 + \sin A)\cos A}$

$= \frac{1 + 1 + 2\sin A}{(1 + \sin A)\cos A} $

$= \frac{2 + 2\sin A}{(1 + \sin A)\cos A}$

$= \frac{2(1 + \sin A)}{(1 + \sin A)\cos A}$

$= \frac{2}{\cos A}$

$= 2\sec A \quad \left[\because \sec A = \frac{1}{\cos A}\right]$

$= \text{RHS}$

$\therefore\text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(iii) $\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \csc \theta$

Solution:

$\text{LHS} = \frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta}$

$= \frac{\frac{\sin \theta}{\cos \theta}}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 – \frac{\sin \theta}{\cos \theta}}$

$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta – \sin \theta}{\cos \theta}}$

$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta – \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta – \sin \theta)}$

$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta – \cos \theta)} – \frac{\cos^2 \theta}{\sin \theta(\sin \theta – \cos \theta)}$

$= \frac{\sin^3 \theta – \cos^3 \theta}{\cos \theta \sin \theta(\sin \theta – \cos \theta)}$

$= \frac{(\sin \theta – \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\cos \theta \sin \theta(\sin \theta – \cos \theta)} $

$= \frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta}{\sin \theta \cos \theta}$

$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \quad [\because \sin^2 \theta + \cos^2 \theta = 1]$

$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$= \frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} + 1$

$= cosec \theta \sec \theta + 1 $

$= 1 + \sec \theta \csc \theta$

$= \text{RHS}$

$\therefore\text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$

Solution:

$\text{LHS} = \frac{1 + \sec A}{\sec A}$

$= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$

$= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}}$

$= \frac{1 + \cos A}{\cos A} \times \cos A$

$= \frac{1 + \cos A}{1}$

 $= \frac{(1 + \cos A)(1 – \cos A)}{1 – \cos A}$ [ Multiplying both numerator and denominator by $(1 – \cos A)$]

$= \frac{1^2 – \cos^2 A}{1 – \cos A} $

$= \frac{1 – \cos^2 A}{1 – \cos A}$

$= \frac{\sin^2 A}{1 – \cos A}$

$= \text{RHS}$

$\therefore\text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = cosecA + \cot A$

$\text{LHS} = \frac{\cos A – \sin A + 1}{\cos A + \sin A – 1}$

$= \frac{\frac{\cos A}{\sin A} – \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} – \frac{1}{\sin A}}$ [Dividing both numerator and denominator by $\sin A$]

$= \frac{\cot A – 1 + \csc A}{\cot A + 1 – \csc A}$

$= \frac{\cot A + cosec A – 1}{\cot A + 1 – cosec A}$

$= \frac{\cot A + cosec A – (cosec^2 A – \cot^2 A)}{\cot A + 1 – cosec A} \quad [\because cosec^2 A – \cot^2 A = 1]$

$= \frac{(\cot A + cosec A) – (cosec A + \cot A)(cosec A – \cot A)}{\cot A + 1 – cosec A} $

$= \frac{(\cot A + cosec A)\{1 – (cosec A – \cot A)\}}{1 – cosec A + \cot A}$

$= \frac{(\cot A + cosec A)(1 – cosec A + \cot A)}{1 – cosec A + \cot A}$

$= \cot A + cosec A$

$= \text{RHS}$

$\therefore\text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$

$\text{LHS} = \sqrt{\frac{1 + \sin A}{1 – \sin A}}$ Multiplying both numerator and denominator by $(1 + \sin A)$ inside the root

$= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 – \sin A)(1 + \sin A)}}$

$= \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} \quad [\because (a+b)(a-b) = a^2 – b^2]$ $= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \quad [\text{Using } \cos^2 A = 1 – \sin^2 A]$

$= \frac{1 + \sin A}{\cos A}$ $= \frac{1}{\cos A} + \frac{\sin A}{\cos A}$

$= \sec A + \tan A \quad \left[\because \frac{1}{\cos A} = \sec A \text{ and } \frac{\sin A}{\cos A} = \tan A\right]$

$= \text{RHS}$ $\dots \text{LHS} = \text{RHS}$ $\text{Hence Proved.}$

(vii) $\frac{\sin \theta – 2\sin^3 \theta}{2\cos^3 \theta – \cos \theta} = \tan \theta$

$\text{LHS} = \frac{\sin \theta – 2\sin^3 \theta}{2\cos^3 \theta – \cos \theta}$

$= \frac{\sin \theta(1 – 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta – 1)}$

$= \tan \theta \left\{ \frac{1 – 2(1 – \cos^2 \theta)}{2\cos^2 \theta – 1} \right\} \quad [\because \sin^2 \theta = 1 – \cos^2 \theta]$

$= \tan \theta \left( \frac{1 – 2 + 2\cos^2 \theta}{2\cos^2 \theta – 1} \right)$

$= \tan \theta \left( \frac{2\cos^2 \theta – 1}{2\cos^2 \theta – 1} \right)$

$= \tan \theta \times 1$

$= \tan \theta$

$= \text{RHS}$

$\dots \text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(viii) $(\sin A + cosec A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$

$\text{LHS} = (\sin A + cosec A)^2 + (\cos A + \sec A)^2$

$= (\sin^2 A + 2\sin A cosec A + cosec ^2 A) + (\cos^2 A + 2\cos A\sec A + \sec^2 A) $

$= \sin^2 A + \cos^2 A + 2\sin A cosec A + 2\cos A\sec A + cosec ^2 A + \sec^2 A$

$= \sin^2 A + \cos^2 A + 2\sin A \cdot \frac{1}{\sin A} + 2\cos A \cdot \frac{1}{\cos A} + cosec ^2 A + \sec^2 A$

$= 1 + 2 + 2 + (\cot^2 A + 1) + (1 + \tan^2 A) $

$= 5 + \cot^2 A + 1 + 1 + \tan^2 A$

$= 7 + \tan^2 A + \cot^2 A$

$= \text{RHS}$

$\dots \text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(ix) $(cosec A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}$

$\text{LHS} = (cosec A – \sin A)(\sec A – \cos A)$

$= \left(\frac{1}{\sin A} – \sin A\right)\left(\frac{1}{\cos A} – \cos A\right)$

$= \left(\frac{1 – \sin^2 A}{\sin A}\right)\left(\frac{1 – \cos^2 A}{\cos A}\right)$

$= \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A} \quad [\because 1 – \sin^2 A = \cos^2 A \text{ and } 1 – \cos^2 A = \sin^2 A]$

$= \cos A \sin A$

$\text{RHS} = \frac{1}{\tan A + \cot A}$

$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$

$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\cos A \sin A}}$

$= \frac{\cos A \sin A}{\sin^2 A + \cos^2 A}$

$= \frac{\cos A \sin A}{1} \quad [\text{Using } \sin^2 A + \cos^2 A = 1]$

$= \cos A \sin A$

$\dots \text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$

$\text{LHS} = \frac{1 + \tan^2 A}{1 + \cot^2 A}$

$= \frac{\sec^2 A}{cosec ^2 A} \quad [\because \sec^2 A = 1 + \tan^2 A, cosec ^2 A = 1 + \cot^2 A]$

$= \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \quad \left[\because \sec A = \frac{1}{\cos A}, cosec A = \frac{1}{\sin A}\right]$

$= \frac{\sin^2 A}{\cos^2 A}$

$= \tan^2 A = \text{RHS}$

$\text{Again, mid term} = \left(\frac{1 – \tan A}{1 – \cot A}\right)^2$

$= (-\tan A)^2$

$= \tan^2 A = \text{RHS}$

$\therefore\text{LHS} = \text{RHS}$

$\text{Hence Proved.}$

Question 6: Supplementary Practice Problems

(i) $tan^4 \theta + \tan^2 \theta = \sec^4 \theta – \sec^2 \theta$

$\text{LHS} = \tan^4 \theta + \tan^2 \theta$

$= \tan^2 \theta(\tan^2 \theta + 1)$

$= (\sec^2 \theta – 1) \times \sec^2 \theta \quad [\because \tan^2 \theta = \sec^2 \theta – 1 \text{ and } \tan^2 \theta + 1 = \sec^2 \theta]$

$= \sec^4 \theta – \sec^2 \theta$

$= \text{RHS}$

$\text{Proved.}$

(ii) $\frac{\cos \theta}{1 – \tan \theta} + \frac{\sin \theta}{1 – \cot \theta} = \sin \theta + \cos \theta$

$\text{LHS} = \frac{\cos \theta}{1 – \tan \theta} + \frac{\sin \theta}{1 – \cot \theta}$

$= \frac{\cos \theta}{1 – \frac{\sin \theta}{\cos \theta}} + \frac{\sin \theta}{1 – \frac{\cos \theta}{\sin \theta}}$

$= \frac{\cos \theta}{\frac{\cos \theta – \sin \theta}{\cos \theta}} + \frac{\sin \theta}{\frac{\sin \theta – \cos \theta}{\sin \theta}}$

$= \frac{\cos^2 \theta}{\cos \theta – \sin \theta} + \frac{\sin^2 \theta}{\sin \theta – \cos \theta}$

$= \frac{\cos^2 \theta}{\cos \theta – \sin \theta} – \frac{\sin^2 \theta}{\cos \theta – \sin \theta}$

$= \frac{\cos^2 \theta – \sin^2 \theta}{\cos \theta – \sin \theta}$

$= \frac{( \cos \theta + \sin \theta )( \cos \theta – \sin \theta )}{\cos \theta – \sin \theta} \quad [\text{Using } a^2 – b^2 = (a+b)(a-b)]$

$= \cos \theta + \sin \theta$

$= \sin \theta + \cos \theta$

$= \text{RHS}$

$\text{Proved.}$

(iii) $\sqrt{\frac{\sec \theta – 1}{\sec \theta + 1}} = cosec \theta – \cot \theta$

$\text{LHS} = \sqrt{\frac{\sec \theta – 1}{\sec \theta + 1}}$

$= \sqrt{\frac{(\sec \theta – 1)(\sec \theta – 1)}{(\sec \theta + 1)(\sec \theta – 1)}}$

$= \sqrt{\frac{(\sec \theta – 1)^2}{\sec^2 \theta – 1}}$

$= \sqrt{\frac{(\sec \theta – 1)^2}{\tan^2 \theta}} \quad [\because \sec^2 \theta – 1 = \tan^2 \theta]$

$= \frac{\sec \theta – 1}{\tan \theta}$

$= \frac{\sec \theta}{\tan \theta} – \frac{1}{\tan \theta}$

$= \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} – \cot \theta \quad \left[\because \sec \theta = \frac{1}{\cos \theta}, \tan \theta = \frac{\sin \theta}{\cos \theta}, \frac{1}{\tan \theta} = \cot \theta\right]$

$= \frac{1}{\sin \theta} – \cot \theta$

$= cosec \theta – \cot \theta \quad \left[\text{Using } \frac{1}{\sin \theta} = cosec \theta\right]$

$= \text{RHS}$

$\text{Proved.}$

(iv) $\cot \theta + \tan \theta = \sec \theta cosec \theta$

$\text{LHS} = \cot \theta + \tan \theta$

$= \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}$

$= \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}$

$= \frac{1}{\sin \theta \cos \theta} \quad [\because \cos^2 \theta + \sin^2 \theta = 1]$

$= \frac{1}{\sin \theta} \times \frac{1}{\cos \theta}$

$= \csc \theta \sec \theta$

$= \sec \theta cosec \theta$

$= \text{RHS}$

$\text{Proved.}$

(v) $\frac{1}{1 + \sin \theta} + \frac{1}{1 – \sin \theta} = 2\sec^2 \theta$

$\text{LHS} = \frac{1}{1 + \sin \theta} + \frac{1}{1 – \sin \theta}$

$= \frac{1 – \sin \theta + 1 + \sin \theta}{(1 + \sin \theta)(1 – \sin \theta)}$

$= \frac{2}{1 – \sin^2 \theta} \quad [\because (a+b)(a-b) = a^2 – b^2]$

$= \frac{2}{\cos^2 \theta} \quad [\text{Using } 1 – \sin^2 \theta = \cos^2 \theta]$

$= 2\sec^2 \theta \quad \left[\because \frac{1}{\cos^2 \theta} = \sec^2 \theta\right]$

$= \text{RHS}$

$\text{Proved.}$

Question 7

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) $\sec A(1 – \sin A)(\sec A + \tan A) = 1$

$\text{LHS} = \sec A(1 – \sin A)(\sec A + \tan A)$

$= \left(\frac{1}{\cos A}\right)(1 – \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$

$= \frac{(1 – \sin A)(1 + \sin A)}{\cos^2 A}$

$= \frac{1 – \sin^2 A}{\cos^2 A} \quad [\because (a-b)(a+b) = a^2 – b^2]$

$= \frac{\cos^2 A}{\cos^2 A} \quad [\text{Using } 1 – \sin^2 A = \cos^2 A]$

$= 1$

$= \text{RHS}$

$\text{Proved.}$

(ii) $\frac{\cot A – \cos A}{\cot A + \cos A} = \frac{cosec A – 1}{cosec A + 1}$

$\text{LHS} = \frac{\cot A – \cos A}{\cot A + \cos A}$

$= \frac{\frac{\cos A}{\sin A} – \cos A}{\frac{\cos A}{\sin A} + \cos A}$

$= \frac{\cos A\left(\frac{1}{\sin A} – 1\right)}{\cos A\left(\frac{1}{\sin A} + 1\right)}$

$= \frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1}$

$= \frac{cosec A – 1}{cosec A + 1} \quad \left[\because \frac{1}{\sin A} = \csc A\right]$

$= \text{RHS}$

$\text{Proved.}$

(iii) $\frac{\cos A}{1 – \tan A} – \frac{\sin^2 A}{\cos A – \sin A} = \sin A + \cos A$

$\text{LHS} = \frac{\cos A}{1 – \tan A} – \frac{\sin^2 A}{\cos A – \sin A}$

$= \frac{\cos A}{1 – \frac{\sin A}{\cos A}} – \frac{\sin^2 A}{\cos A – \sin A}$

$= \frac{\cos A}{\frac{\cos A – \sin A}{\cos A}} – \frac{\sin^2 A}{\cos A – \sin A}$

$= \frac{\cos^2 A}{\cos A – \sin A} – \frac{\sin^2 A}{\cos A – \sin A}$

$= \frac{\cos^2 A – \sin^2 A}{\cos A – \sin A}$

$= \frac{(\cos A – \sin A)(\cos A + \sin A)}{\cos A – \sin A} \quad [\because a^2 – b^2 = (a-b)(a+b)]$

$= \cos A + \sin A$

$= \sin A + \cos A$

$= \text{RHS}$

$\text{Proved.}$

(iv) $\frac{cosec \theta + \cot \theta}{cosec \theta – \cot \theta} = 1 + 2\cot^2 \theta + 2cosec \theta \cot \theta$

$\text{LHS} = \frac{cosec \theta + \cot \theta}{cosec \theta – \cot \theta}$ Multiplying both numerator and denominator by $(\csc \theta + \cot \theta)$:

$= \frac{(cosec \theta + \cot \theta)(cosec \theta + \cot \theta)}{(cosec \theta – \cot \theta)(cosec \theta + \cot \theta)}$

$= \frac{(cosec \theta + \cot \theta)^2}{cosec ^2 \theta – \cot^2 \theta}$

$= \frac{cosec ^2 \theta + \cot^2 \theta + 2cosec \theta \cot \theta}{1} \quad [\because cosec ^2 \theta – \cot^2 \theta = 1]$

$= (1 + \cot^2 \theta) + \cot^2 \theta + 2cosec \theta \cot \theta \quad [\text{Using } cosec ^2 \theta = 1 + \cot^2 \theta]$

$= 1 + 2\cot^2 \theta + 2cosec \theta \cot \theta$

$= \text{RHS}$ $\text{Proved.}$

(v) $\sqrt{\frac{1 + \cos \beta}{1 – \cos \beta}} = cosec \beta + \cot \beta$

$\text{LHS} = \sqrt{\frac{1 + \cos \beta}{1 – \cos \beta}}$ [Multiplying both numerator and denominator by $(1 + \cos \beta)$ inside the root]

$= \sqrt{\frac{(1 + \cos \beta)(1 + \cos \beta)}{(1 – \cos \beta)(1 + \cos \beta)}}$

$= \sqrt{\frac{(1 + \cos \beta)^2}{1 – \cos^2 \beta}}$

$= \sqrt{\frac{(1 + \cos \beta)^2}{\sin^2 \beta}} \quad [\text{Using } 1 – \cos^2 \beta = \sin^2 \beta]$

$= \frac{1 + \cos \beta}{\sin \beta}$

$= \frac{1}{\sin \beta} + \frac{\cos \beta}{\sin \beta}$

$= cosec \beta + \cot \beta \quad \left[\because \frac{1}{\sin \beta} = cosec \beta \text{ and } \frac{\cos \beta}{\sin \beta} = \cot \beta\right]$

$= \text{RHS}$ $\text{Proved.}$

(vi) $\sqrt{\frac{1 + \cos \theta}{1 – \cos \theta}} + \sqrt{\frac{1 – \cos \theta}{1 + \cos \theta}} = 2\csc \theta$

$\text{LHS} = \sqrt{\frac{1 + \cos \theta}{1 – \cos \theta}} + \sqrt{\frac{1 – \cos \theta}{1 + \cos \theta}}$

$= \frac{(\sqrt{1 + \cos \theta})^2 + (\sqrt{1 – \cos \theta})^2}{\sqrt{(1 – \cos \theta)(1 + \cos \theta)}}$

$= \frac{(1 + \cos \theta) + (1 – \cos \theta)}{\sqrt{1 – \cos^2 \theta}}$

$= \frac{2}{\sqrt{\sin^2 \theta}} \quad [\text{Using } 1 – \cos^2 \theta = \sin^2 \theta]$

$= \frac{2}{\sin \theta}$

$= 2\csc \theta \quad \left[\because \frac{1}{\sin \theta} = \csc \theta\right]$

$= \text{RHS}$

$\text{Proved.}$

(vii) $\sqrt{\frac{\sec \theta + 1}{\sec \theta – 1}} = \cot \theta + \csc \theta$

$\text{LHS} = \sqrt{\frac{\sec \theta + 1}{\sec \theta – 1}}$ [Multiplying both numerator and denominator by $(\sec \theta + 1)$ inside the root]

$= \sqrt{\frac{(\sec \theta + 1)(\sec \theta + 1)}{(\sec \theta – 1)(\sec \theta + 1)}}$

$= \sqrt{\frac{(\sec \theta + 1)^2}{\sec^2 \theta – 1}}$

$= \sqrt{\frac{(\sec \theta + 1)^2}{\tan^2 \theta}} \quad [\text{Using } \sec^2 \theta – 1 = \tan^2 \theta]$

$= \frac{\sec \theta + 1}{\tan \theta}$

$= \frac{\sec \theta}{\tan \theta} + \frac{1}{\tan \theta}$

$= \frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} + \cot \theta$ $= \frac{1}{\sin \theta} + \cot \theta$

$= cosec \theta + \cot \theta \quad \left[\because \frac{1}{\sin \theta} = cosec \theta\right]$

$= \cot \theta + cosec \theta$ $= \text{RHS}$ $\text{Proved.}$

(viii) $\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} – 2$

$\text{LHS} = \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta}$

$= \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta}$

$= \frac{(\sin^2 \theta + \cos^2 \theta)^2 – 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \quad [\because a^2 + b^2 = (a+b)^2 – 2ab]$

$= \frac{(1)^2 – 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \quad [\text{Using } \sin^2 \theta + \cos^2 \theta = 1]$

$= \frac{1 – 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$

$= \frac{1}{\sin^2 \theta \cos^2 \theta} – \frac{2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$

$= \frac{1}{\sin^2 \theta \cos^2 \theta} – 2$

$= \text{RHS}$

$\text{Proved.}$

(ix) $\sin \theta \cdot \cos(90^\circ – \theta) + \cos \theta \cdot \sin(90^\circ – \theta) = 1$

$\text{LHS} = \sin \theta \cdot \cos(90^\circ – \theta) + \cos \theta \cdot \sin(90^\circ – \theta)$

$= \sin \theta \cdot \sin \theta + \cos \theta \cdot \cos \theta \quad [\because \cos(90^\circ – \theta) = \sin \theta \text{ and } \sin(90^\circ – \theta) = \cos \theta]$

$= \sin^2 \theta + \cos^2 \theta$

$= 1 \quad [\text{Using } \sin^2 \theta + \cos^2 \theta = 1]$

$= \text{RHS}$

$\text{Proved.}$

(x) $\frac{1}{\sec \theta – \tan \theta} – \frac{1}{\cos \theta} = \frac{1}{\cos \theta} – \frac{1}{\sec \theta + \tan \theta}$

Let us rearrange the terms, we get:

$\frac{1}{\sec \theta – \tan \theta} + \frac{1}{\sec \theta + \tan \theta} = \frac{1}{\cos \theta} + \frac{1}{\cos \theta}$

$\text{New LHS} = \frac{1}{\sec \theta – \tan \theta} + \frac{1}{\sec \theta + \tan \theta}$

$= \frac{(\sec \theta + \tan \theta) + (\sec \theta – \tan \theta)}{(\sec \theta – \tan \theta)(\sec \theta + \tan \theta)}$

$= \frac{2\sec \theta}{\sec^2 \theta – \tan^2 \theta}$

$= \frac{2\sec \theta}{1} \quad [\text{Using } \sec^2 \theta – \tan^2 \theta = 1]$

$= 2\sec \theta$

$= \frac{2}{\cos \theta}$

$\text{New RHS} = \frac{1}{\cos \theta} + \frac{1}{\cos \theta}$

$= \frac{2}{\cos \theta}$

$\because \text{New LHS} = \text{New RHS}$

$\therefore \frac{1}{\sec \theta – \tan \theta} – \frac{1}{\cos \theta} = \frac{1}{\cos \theta} – \frac{1}{\sec \theta + \tan \theta}$

$\text{Proved.}$

(xi) $\sec^2 \alpha \csc^2 \alpha = \tan^2 \alpha + \cot^2 \alpha + 2$

Sol:

$\text{RHS} = \tan^2 \alpha + \cot^2 \alpha + 2$

$= (\tan^2 \alpha + 1) + (\cot^2 \alpha + 1)$

$= \sec^2 \alpha + cosec^2 \alpha \quad [\because 1 + \tan^2 \alpha = \sec^2 \alpha \text{ and } 1 + \cot^2 \alpha = \cosec ^2 \alpha]$

$= \frac{1}{\cos^2 \alpha} + \frac{1}{\sin^2 \alpha}$

$= \frac{\sin^2 \alpha + \cos^2 \alpha}{\cos^2 \alpha \sin^2 \alpha}$

$= \frac{1}{\cos^2 \alpha \sin^2 \alpha} \quad [\text{Using } \sin^2 \alpha + \cos^2 \alpha = 1]$

$= \left(\frac{1}{\cos^2 \alpha}\right)\left(\frac{1}{\sin^2 \alpha}\right)$

$= \sec^2 \alpha cosec ^2 \alpha$

$= \text{LHS}$

$\text{Proved.}$

(xii) $\tan^2 A – \tan^2 B = \frac{\sin^2 A – \sin^2 B}{\cos^2 A \cos^2 B}$

$\text{LHS} = \tan^2 A – \tan^2 B$

$= \frac{\sin^2 A}{\cos^2 A} – \frac{\sin^2 B}{\cos^2 B}$

$= \frac{\sin^2 A \cos^2 B – \sin^2 B \cos^2 A}{\cos^2 A \cos^2 B}$

$= \frac{\sin^2 A(1 – \sin^2 B) – \sin^2 B(1 – \sin^2 A)}{\cos^2 A \cos^2 B} \quad [\text{Using } \cos^2 \theta = 1 – \sin^2 \theta]$

$= \frac{\sin^2 A – \sin^2 A \sin^2 B – \sin^2 B + \sin^2 B \sin^2 A}{\cos^2 A \cos^2 B}$

$= \frac{\sin^2 A – \sin^2 B}{\cos^2 A \cos^2 B}$

$= \text{RHS}$

$\text{Proved.}$

Question 8

(i) If $\cos \theta + \sin \theta = \sqrt{2}\cos \theta$, show that $\cos \theta – \sin \theta = \sqrt{2}\sin \theta$

$\text{Given: } \cos \theta + \sin \theta = \sqrt{2}\cos \theta$

$\Rightarrow \sin \theta = \sqrt{2}\cos \theta – \cos \theta$

$\Rightarrow \sin \theta = (\sqrt{2} – 1)\cos \theta$

Multiplying both sides by $(\sqrt{2} + 1)$:

$\Rightarrow (\sqrt{2} + 1)\sin \theta = (\sqrt{2} + 1)(\sqrt{2} – 1)\cos \theta$

$\Rightarrow \sqrt{2}\sin \theta + \sin \theta = ((\sqrt{2})^2 – 1^2)\cos \theta \quad [\because (a+b)(a-b) = a^2 – b^2]$

$\Rightarrow \sqrt{2}\sin \theta + \sin \theta = (2 – 1)\cos \theta$

$\Rightarrow \sqrt{2}\sin \theta + \sin \theta = \cos \theta$ $\therefore \cos \theta – \sin \theta = \sqrt{2}\sin \theta$

$\text{Hence Proved.}$

(ii) If $\sin \theta + \sin^2 \theta = 1$, prove that $\cos^2 \theta + \cos^4 \theta = 1$

$\text{Given: } \sin \theta + \sin^2 \theta = 1$

$\Rightarrow \sin \theta = 1 – \sin^2 \theta$

$\Rightarrow \sin \theta = \cos^2 \theta \quad [\because 1 – \sin^2 \theta = \cos^2 \theta]$

Squaring both sides: $\Rightarrow \sin^2 \theta = (\cos^2 \theta)^2$

$\Rightarrow \sin^2 \theta = \cos^4 \theta$ $\Rightarrow 1 – \cos^2 \theta = \cos^4 \theta \quad [\because \sin^2 \theta = 1 – \cos^2 \theta]$

$\therefore \cos^2 \theta + \cos^4 \theta = 1$ $\text{Hence Proved.}$

Question 9

$\sin \theta$ can be expressed as:

(P) $\frac{1}{cosec \theta}$

(Q) $\frac{1}{\sqrt{1 + \tan^2 \theta}}$

(R) $\frac{1}{\sqrt{1 + \cot^2 \theta}}$

(S) $\frac{1}{1 + \cot^2 \theta}$

Choose the correct option:

(a) Both P and S are true

(b) Both P and Q are true

(c) Both P and R are true

(d) Both R and S are true

Ans: (c) Both P and R are true

Justification:

Statement P: By definition, $\sin \theta = \frac{1}{cosec \theta}$. This is true.

Statement R: We know that $cosec^2 \theta = 1 + \cot^2 \theta \Rightarrow \csc \theta = \sqrt{1 + \cot^2 \theta}$.

Therefore, $\sin \theta = \frac{1}{cosec \theta} = \frac{1}{\sqrt{1 + \cot^2 \theta}}$. This is true.

Therefore, the correct option is (c) Both P and R are true.

Question 10

Match the column I and column II and then choose the correct option from the given alternatives.

(a) P-2, Q-3, R-1

(b) P-3, Q-2, R-1

(c) P-2, Q-1, R-3

(d) P-3, Q-1, R-2

Ans: (a) P-2, Q-3, R-1

Justification:

Match for P: $\frac{1}{\cos^2 \theta} = \sec^2 \theta = 1 + \tan^2 \theta$. So, P matches with 2.

Match for Q: $1 – \cos^2 \theta = \sin^2 \theta = \frac{1}{cosec ^2 \theta}$. So, Q matches with 3.

Match for R: $\frac{1}{\sin^2 \theta} = cosec ^2 \theta = 1 + \cot^2 \theta$. So, R matches with 1.

Therefore, the correct option is (a).