Solution

We know that,

$cosec^{2} A=1+\cot ^{2} A$

$\dfrac{1}{cosec^{2} A}=\dfrac{1}{1+\cot ^{2} A}$

$\sin ^{2} A=\dfrac{1}{1+\cot ^{2} A}$

$\sin A= \pm \dfrac{1}{\sqrt{1+\cot ^{2} A}}$

$\sqrt{1+\cot ^{2} A}$ will always be positive as we are adding two positive quantities.

Therefore, $\sin A=\dfrac{1}{\sqrt{1+\cot ^{2} A}}$

We know that, $\tan A=\dfrac{\sin A}{\cos A}$

However, $\cot A=\dfrac{\cos A}{\sin A}$

Therefore, $\tan A=\dfrac{1}{\cot A}$

Also, $\sec ^{2} A=1+\tan ^{2} A$

$=1+\dfrac{1}{\cot ^{2} A}$

$=\dfrac{\cot ^{2} A+1}{\cot ^{2} A}$

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

Solution

We know that,

$\cos A=\dfrac{1}{\sec A}$

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

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

$ \begin{aligned} \sin A & =\sqrt{1-(\dfrac{1}{\sec A})^{2}} \\ & =\sqrt{\dfrac{\sec ^{2} A-1}{\sec ^{2} A}}=\dfrac{\sqrt{\sec ^{2} A-1}}{\sec A} \end{aligned} $

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

$\tan ^{2} A=\sec ^{2} A-1$

$ \begin{aligned} \tan A & =\sqrt{\sec ^{2} A-1} \\ \cot A & =\dfrac{\cos A}{\sin A}=\dfrac{\dfrac{1}{\sec A}}{\dfrac{\sqrt{\sec ^{2} A-1}}{\sec A}} \\ & =\dfrac{1}{\sqrt{\sec ^{2} A-1}} \end{aligned} $

$cosec A=\dfrac{1}{\sin A}=\dfrac{\sec A}{\sqrt{\sec ^{2} A-1}}$

Solution

(i) $9 \sec ^{2} A-9 \tan ^{2} A$

$=9(\sec ^{2} A-\tan ^{2} A)$

$=9$ (1) [As $.\sec ^{2} A-\tan ^{2} A=1]$

$=9$

Hence, alternative (B) is correct.

(ii)

$ \begin{aligned} & (1+\tan \theta+\sec \theta)(1+\cot \theta-cosec \theta) \\ & =(1+\dfrac{\sin \theta}{\cos \theta}+\dfrac{1}{\cos \theta})(1+\dfrac{\cos \theta}{\sin \theta}-\dfrac{1}{\sin \theta}) \\ & =(\dfrac{\cos \theta+\sin \theta+1}{\cos \theta})(\dfrac{\sin \theta+\cos \theta-1}{\sin \theta}) \\ & =\dfrac{(\sin \theta+\cos \theta)^{2}-(1)^{2}}{\sin \theta \cos \theta} \\ & =\dfrac{\sin \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\ & =\dfrac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\ & =\dfrac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2 \end{aligned} $

Hence, alternative (C) is correct.

(iii) $(\sec A+\tan A)(1-\sin A)$

$ \begin{aligned} & =(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A})(1-\sin A) \\ & =(\dfrac{1+\sin A}{\cos A})(1-\sin A) \\ & =\dfrac{1-\sin ^{2} A}{\cos A}=\dfrac{\cos ^{2} A}{\cos A} \\ & =\cos A \end{aligned} $

Hence, alternative (D) is correct.

(iv)

$ \dfrac{1+\tan ^{2} A}{1+\cot ^{2} A}=\dfrac{1+\dfrac{\sin ^{2} A}{\cos ^{2} A}}{1+\dfrac{\cos ^{2} A}{\sin ^{2} A}} $

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

$=\dfrac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A$

Hence, alternative (D) is correct.

Solution

(i)

$(cosec \theta-\cot \theta)^{2}=\dfrac{1-\cos \theta}{1+\cos \theta}$

L.H.S. $=(cosec \theta-\cot \theta)^{2}$

$=(\dfrac{1}{\sin \theta}-\dfrac{\cos \theta}{\sin \theta})^{2}$

$=\dfrac{(1-\cos \theta)^{2}}{(\sin \theta)^{2}}=\dfrac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$

$=\dfrac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}=\dfrac{(1-\cos \theta)^{2}}{(1-\cos \theta)(1+\cos \theta)}=\dfrac{1-\cos \theta}{1+\cos \theta}$

$=$ R.H.S.

(ii)

$\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2 \sec A$

L.H.S. $=\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$

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

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

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

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

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

$=$ R.H.S.

(iii) $\dfrac{\tan \theta}{1-\cot \theta}+\dfrac{\cot \theta}{1-\tan \theta}=1+\sec \theta cosec \theta$

$ \begin{aligned} \text{ L.H.S. } & =\dfrac{\tan \theta}{1-\cot \theta}+\dfrac{\cot \theta}{1-\tan \theta} \\ & =\dfrac{\dfrac{\sin \theta}{\cos \theta}}{1-\dfrac{\cos \theta}{\sin \theta}}+\dfrac{\dfrac{\cos \theta}{\sin \theta}}{1-\dfrac{\sin \theta}{\cos \theta}} \\ & =\dfrac{\dfrac{\cos \theta}{\cos \theta - \sin \theta}}{\cos \theta} by \dfrac{\dfrac{\cos \theta}{\sin \theta}{\cos \theta - \sin \theta}}{\cos \theta}\\ = & \dfrac{\sin ^{2} \theta}{\cos \theta(\sin ^{2} \theta-\cos \theta)}-\dfrac{\cos ^{2} \theta}{\sin ^{2} \theta(\sin \theta-\cos \theta)} \end{aligned} $

(iv)

By LHS

$\begin{aligned} & =\dfrac{\dfrac{1}{1}+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}} \\ \\ & =\frac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\ \\ & =\cos A+1 \\ & =1+\cos A \\ \end{aligned}$

By RHS

$=\dfrac{1^2 -\cos ^2 A}{1-\cos A}$

$ \begin{array}{l} \because a^2-b^2=(a+b)(a-b) \\ \sin ^2 A+\cos ^2 A=1 \\ \sin ^2 A=1-\cos ^2 A \\ \end{array}$

$=\dfrac{(1+\cos A) (1-\cos A)} {(1-\cos A)}$

$1+\cos A$

LHS $=$ RHS

(v)

By LHS

$\text { Divide each term by } \sin \mathrm{A}$

$\begin{aligned} & =\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\sin A 1}{\sin A}+\dfrac{1}{\sin A}} {\dfrac{\cos A}{\sin A}+\dfrac{\sin A 1}{\sin A}-\dfrac{1}{\sin A}} \\ \\ & =\dfrac{\cot A-1+\operatorname{cosec} A}{\cot A+1-\operatorname{cosec} A} \\ \\ & =\frac{\cot A+\operatorname{cosec} A-\left(\operatorname{cosec}^2 A-\cot ^2 A\right)}{\cot A+1-\operatorname{cosec} A} \\ \\ & =\dfrac{\cot A+\operatorname{cosec} A-(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)}{\cot A+1-\operatorname{cosec} A} \\ \\ & =\dfrac{\cot A+\operatorname{cosec} A[1-(\operatorname{cosec} A+\cot A)]}{\cot A+1-\operatorname{cosec} A} \\ \\ & =\dfrac{\cot A+\operatorname{cosec} A(1-\operatorname{cosec} A+\cot A)}{\cot A+1-\operatorname{cosec} A} \\ \\ & =\cot A+\operatorname{cosec} A \\ \\ & =\operatorname{cosec} A+\cot A &\end{aligned}$

= RHS

(vi)

By LHS

$\begin{aligned} & =\sqrt{\dfrac{(1+\sin A) \times(1+\sin A)}{(1-\sin A) \times(1+\sin A)}} \\ \\ & =\sqrt{\dfrac{(1+\sin A)^2}{1^2-\sin ^2 A}} \\ \\ & =\sqrt{\dfrac{(1+\sin A)^2}{\cos ^2 A}} \\ \\ & =\dfrac{\sqrt{(1+\sin A)^2}}{\sqrt{\cos^2 A}} \\ \\ & =\dfrac{1+\sin A}{\cos A} \\ \\ & \because (a-b)(a+b)=a^2-b^2 \\ \\ & \sin ^2 A+\cos ^2 A=1 \\ \\ & \cos ^2 A=1-\sin ^2 A \\ \\ & =\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \\ \\ & =\sec A+\tan A \end{aligned}$

= RHS Hence proved

(vii)

By LHS

$\begin{aligned} &= \dfrac{\sin \theta\left(1-2 \sin ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-1\right)} \\ \\ &=\dfrac{\sin \theta\left(\sin ^2 \theta+\cos ^2 \theta-2 \sin ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-\left(\sin ^2 \theta+\cos ^2 \theta\right)\right.} \\ \\ &=\dfrac{\sin \theta\left(-\sin ^2 \theta+\cos ^2 \theta\right)}{\cos \theta\left(2 \cos ^2 \theta-\sin ^2 \theta-\cos ^2 \theta\right)} \\ \\ &\because \sin ^2 \theta+\cos ^2 \theta=1 , \sin ^2 \theta-2 \sin ^2 \theta = -\sin ^2 \theta, 2 \cos ^2 \theta-\cos ^2 \theta =\cos ^2 \theta\\ \\ &=\dfrac{\sin \theta\left(\cos ^2 \theta-\sin ^2 \theta\right)}{\cos \theta\left(\cos ^2 \theta-\sin ^2 \theta\right)} \end{aligned}$

$\begin{aligned} =\frac{\sin \theta}{\cos \theta}=\tan \theta \end{aligned}$ = RHS Hence proved

(viii)

By LHS

$\begin{aligned} & =\sin ^2 A+\operatorname{cosec}^2 A+2 \sin A \operatorname{cosec} A+\cos ^2 A+\sec ^2 A+2 \cos A \sec A \\ \\ & \because (a+b)^2=a^2+b^2+2 a b \\ & \sec ^2 A=1+\tan ^2 \theta \\ & \operatorname{cosec}^2 A=1+\cot ^2 A \\ \\ & =\sin ^2 A+\cos ^2 A+\operatorname{cosec}^2 A+\sec ^2 A+2 \sin A \operatorname{cosec} A+2 \cos A \sec A \\ \\ & =1+1+\cot ^2+1+\tan ^2 A+2 \sin A \frac{1}{\sin A}+2 \cos A \times \frac{1}{\cos A} \\ \\ & =2+\cot ^2 A+1+\tan ^2 A+2+2 \\ \\ & =7+\cot ^2 A+\tan ^2 A \end{aligned}$

= RHS Hence proved

(ix)

LHS

$\begin{aligned} & \left(\dfrac{1}{\sin A}-\dfrac{\sin A}{1}\right)\left(\dfrac{1}{\cos A}-\dfrac{\cos A}{1}\right) \\ \\ &=\left(\dfrac{1-\sin ^2 A}{\sin A}\right)\left(\dfrac{1-\cos ^2 A}{\cos A}\right) \\ \\ &=\left(\dfrac{\cos ^2 A}{\sin A}\right)\left(\dfrac{\sin ^2 A}{\cos A}\right) \\ \\ &= \dfrac{\cos A \cos A}{\sin A} \times \dfrac{\sin A \sin A}{\cos A} \\ \\ &=\cos A \sin A &\end{aligned}$

RHS

$\begin{aligned} & =\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}} \\ \\ & =\dfrac{1}{\dfrac{\sin ^2 A+\cos ^2 A}{\cos A \sin A}} \\ \\ & \because \sin^2 A+\cos ^2 A=1 \\ & \hspace{.6cm}\sin^2 A=1-\cos ^2 A \quad\quad \\ & =\dfrac{1}{\dfrac{1}{\cos A \sin A}} \\ \\ & =1 \times \dfrac{\cos A \sin A}{1} \\ \\ & =\cos A \sin A &\end{aligned}$

LHS = RHS Hence proved

(x)

LHS

$\begin{aligned} & =\dfrac{\sec ^2 A}{\operatorname{cosec}^2 A} \\ & =\dfrac{\dfrac{1}{\cos ^2 A}}{\dfrac{1}{\sin ^2 A}} \\ & =\dfrac{1}{\cos ^2 A} \times \dfrac{\sin ^2 A}{1} \\ & =\dfrac{\sin ^2 A}{\cos ^2 A}=\tan ^2 A \\ & =\left(\frac{1-\tan A}{\frac{1}{1}-\frac{1}{\tan A}}\right)^2 \\ & =\left[\frac{1-\tan A}{\frac{\tan A-1}{\tan A}}\right]^2 \\ & =\left[\frac{1-\tan A}{\frac{-(1-\tan A)}{\tan A}}\right]^2 \\ & =\left[(1-\tan A) \times \frac{(-\tan A)}{(1-\tan A)}\right]^2 \\ & =[-\tan A]^2 \\ & =\tan ^2 A \\ \end{aligned}$

LHS = RHS