Prove the following trigonometric identities. (1 ^2 - ^2 +1 cosec…

Question

Prove the following trigonometric identities.
$\Big(\frac{1}{\sec^2\theta-\cos^2\theta}+\frac{1}{\text{cosec}^2\theta-\sin^2\theta}\Big)\sin^2\theta\cos^2\theta=\frac{1-\sin^2\theta\cos^2\theta}{2+\sin^2\theta\cos^2\theta}$

✓

Answer

$\text{L.H.S}=\Big(\frac{1}{\sec^2\theta-\cos^2\theta}+\frac{1}{\text{cosec}^2\theta-\sin^2\theta}\Big)\sin^2\theta\cos^2\theta$
$\Rightarrow \bigg[\frac{1}{\frac{1}{\cos^2\theta}-\cos^2\theta}+\frac{1}{\text{cosec}^2\theta-\sin^2\theta}\bigg]\sin^2\theta\cos^2\theta$
$=\Bigg[\frac{1}{\frac{1-\cos^4\theta}{\cos^2\theta}}+\frac{1}{\frac{1-\sin^4\theta}{\sin^2\theta}}\Bigg]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{1-\cos^4\theta}+\frac{\sin^2\theta}{1-\sin^4\theta}\Big]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{\cos^2\theta+\sin^2\theta-\cos^4\theta}+\frac{\sin^2}{\cos^2\theta+\sin^2\theta-\sin^4\theta}\Big]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{\cos^2\theta(1-\cos^2\theta)+\sin^2\theta}+\frac{\sin^2}{\sin^2\theta(1-\sin^2\theta)+\cos^2\theta}\Big]\sin^2\theta\cos^2\theta$
$=\Big[\frac{\cos^2\theta}{\cos^2\theta\sin^2\theta+\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta\cos^2\theta+\cos^2\theta}\Big]\sin^2\theta\cos^2\theta$
$=\bigg[\frac{\cos^2\theta}{\sin^2\theta(\cos^2\theta+1)}+\frac{\sin^2\theta}{\cos^2\theta(\sin^2\theta+1)}\bigg]\sin^2\theta\cos^2\theta$
$=\bigg[\frac{\cos^4\theta(1+\sin^2\theta)+\sin^4\theta(1+\cos^2\theta)}{\sin^2\theta\cos^2\theta(1+\cos^2\theta)(1+\sin^2\theta)}\bigg]\sin^2\theta\cos^2\theta$
$=\frac{\cos^4\theta(1+\sin^2\theta)+\sin^4\theta(1+\cos^2\theta)}{(1+\cos^2\theta)(1+\sin^2\theta)}$
$=\frac{\cos^4\theta+\cos^4\theta\sin^2\theta+\sin^4\theta+\sin^4\theta\cos^2\theta}{1+\sin^2\theta+\cos^2\theta+\cos^2\theta\sin^2\theta}$
$=\frac{1-2\sin^2\theta\cos^2\theta+\sin^2\theta\cos^2\theta(\cos^2\theta+\sin^2\theta)}{1+1+\cos^2\theta\sin^2\theta}$
$(\because\ \cos^2\theta+\sin^2\theta=1)$
$=\frac{1-\sin^2\theta\cos^2\theta}{2+\sin^2\theta\cos^2\theta}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$