# Prove the following identities:$\left(\frac{1}{\sec ^{2} \theta-\cos ^{2} \theta}+\frac{1}{\operatorname{cosec}^{2} \theta-\sin ^{2} \theta}\right) \sin ^{2} \theta \cos ^{2} \theta=\frac{1-\sin ^{2} \theta \cos ^{2} \theta}{2+\sin ^{2} \theta \cos ^{2} \theta}$

To do:

We have to prove that $\left(\frac{1}{\sec ^{2} \theta-\cos ^{2} \theta}+\frac{1}{\operatorname{cosec}^{2} \theta-\sin ^{2} \theta}\right) \sin ^{2} \theta \cos ^{2} \theta=\frac{1-\sin ^{2} \theta \cos ^{2} \theta}{2+\sin ^{2} \theta \cos ^{2} \theta}$.

Solution:

We know that,

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

$\operatorname{cosec}^2 A-\cot^2 A=1$

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

$\cot A=\frac{\cos A}{\sin A}$

$\tan A=\frac{\sin A}{\cos A}$

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

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

Therefore,

$\left(\frac{1}{\sec ^{2} \theta-\cos ^{2} \theta}+\frac{1}{\operatorname{cosec}^{2} \theta-\sin ^{2} \theta}\right) \sin ^{2} \theta \cos ^{2} \theta=\left[\frac{1}{\frac{1}{\cos ^{2} \theta}-\cos ^{2} \theta}+\frac{1}{\frac{1}{\sin ^{2} \theta}-\sin ^{2} \theta}\right] \sin ^{2} \theta \cos ^{2} \theta$

$=\left[\frac{1}{\frac{1-\cos ^{4} \theta}{\cos ^{2} \theta}}+\frac{1}{\frac{1-\sin ^{4} \theta}{\sin ^{2} \theta}}\right] \sin ^{2} \theta \cos ^{2} \theta$

$=\left[\frac{\cos ^{2} \theta}{1-\cos ^{4} \theta}+\frac{\sin ^{2} \theta}{1-\sin ^{4} \theta}\right] \sin ^{2} \theta \cos ^{2} \theta$
$=\left[\frac{\cos ^{2} \theta}{\cos ^{2} \theta+\sin ^{2} \theta-\cos ^{4} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta+\sin ^{2} \theta-\sin ^{4} \theta}\right] \sin ^{2} \theta \cos ^{2} \theta$

$=\left[\frac{\cos ^{2} \theta}{\cos ^{2} \theta\left(1-\cos ^{2} \theta\right)+\sin ^{2} \theta}+\frac{\sin ^{2} \theta}{\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)+\cos ^{2} \theta}\right] \sin ^{2} \theta \cos ^{2} \theta$

$=\left[\frac{\cos ^{2} \theta}{\cos ^{2} \theta\left(\sin ^{2} \theta\right)+\sin ^{2} \theta}+\frac{\sin ^{2} \theta}{\sin ^{2} \theta\left(\cos ^{2} \theta\right)+\cos ^{2} \theta}\right] \sin ^{2} \theta \cos ^{2} \theta$

$=\left[\frac{\cos ^{4} \theta\left(1+\sin ^{2} \theta\right)+\sin ^{4} \theta\left(1+\cos ^{2} \theta\right)}{\sin ^{2} \theta \cos ^{2} \theta\left(1+\cos ^{2} \theta\right)\left(1+\sin ^{2} \theta\right)}\right] \sin ^{2} \theta \cos ^{2} \theta$

$=\frac{\cos ^{4} \theta\left(1+\sin ^{2} \theta\right)+\sin ^{4} \theta\left(1+\cos ^{2} \theta\right)}{\left(1+\cos ^{2} \theta\right)\left(1+\sin ^{2} \theta\right)}$

$=\frac{\cos ^{4} \theta+\cos ^{4} \theta \sin ^{2} \theta+\sin ^{4} \theta+\sin ^{4} \theta \cos ^{2} \theta}{1+\sin ^{2} \theta+\cos ^{2} \theta+\cos ^{2} \theta \sin ^{2} \theta}$

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

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

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

Hence proved.

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Updated on: 10-Oct-2022

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