Prove the following identities: 1 ( ^2 - ^2 ) +1 (cosec^2 - ^2 ) ( ^2 ^2 ) =1- ^2 ^2 2+ ^2 ^2

LHS= 1 ( ^2 - ^2 ) +1 (cosec^2 - ^2 ) ( ^2 ^2 ) = 1 1 ^2 - ^2 +1 1 ^2 - ^2 ^2 ^2 = ^2 ^2 ^2 (1- ^4 ) + ^4 ^2 ^2 1- ^4 = ^2 ^4 (1- ^2 ) (1- ^2 ) + ^4 ^2 1- ^2 (1+ ^2 ) = ^4 (1+ ^2 ) + ^4 1+ ^2 = ^4 + ^4 ^2 + ^4 + ^4 ^2 (1+ ^2 ) (1+ ^2 ) = ^4 + ^4 + ^2 ^2 ( ^2 + ^2 ) 1+ ^2 + ^2 + ^2 + ^2 [ ^2+b^2=(a+b)^2-2ab ] =1- ^2 ^2 2+ ^2 ^2 =R.H.S R.H.S.=L.H.S.

Prove the following identities: 1 ( ^2 - ^2 ) +1 (cosec^2 - ^2 ) …

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Question

Prove the following identities:
$\Bigg\{\frac{1}{\big(\sec^2\theta-\cos^2\theta\big)}+\frac{1}{\big(\text{cosec}^2\theta-\sin^2\theta\big)}\Bigg\}\big(\sin^2\theta\cos^2\theta\big)$
$=\frac{1-\sin^2\theta\cos^2\theta}{2+\sin^2\theta\cos^2\theta}$

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Answer

$\text{LHS}=\Bigg\{\frac{1}{\big(\sec^2\theta-\cos^2\theta\big)}+\frac{1}{\big(\text{cosec}^2\theta-\sin^2\theta\big)}\Bigg\}\big(\sin^2\theta\cos^2\theta\big)$
$=\Bigg\{\frac{1}{\frac{1}{\cos^2\theta}-\cos^2\theta}+\frac{1}{\frac{1}{\sin^2\theta}-\sin^2\theta}\Bigg\}\times\sin^2\theta\cos^2\theta$
$=\Bigg\{\frac{\sin^2\theta\cos^2\times\cos^2\theta}{\big(1-\cos^4\theta\big)}+\frac{\sin^4\theta\cos^2\theta\sin^2\theta}{1-\sin^4\theta}\Bigg\}$
$=\Bigg\{\frac{\sin^2\theta\times\cos^4\theta}{\big(1-\cos^2\theta\big)\big(1-\cos^2\theta\big)}+\frac{\sin^4\theta\cos^2\theta}{1-\sin^2\theta\big(1+\sin^2\theta\big)}\Bigg\}$
$=\Bigg\{\frac{\cos^4\theta}{\big(1+\cos^2\theta\big)}+\frac{\sin^4\theta}{1+\sin^2\theta}\Bigg\}$
$=\frac{\cos^4\theta+\cos^4\theta\sin^2\theta+\sin^4\theta+\sin^4\theta\cos^2\theta}{\big(1+\cos^2\theta\big)\big(1+\sin^2\theta\big)}$
$=\frac{\cos^4\theta+\sin^4\theta+\cos^2\theta\sin^2\theta\big(\cos^2\theta+\sin^2\theta\big)}{1+\sin^2\theta+\cos^2\theta+\cos^2\theta+\sin^2\theta}$ $\Big[\because\text{a}^2+\text{b}^2=(\text{a}+\text{b})^2-2\text{ab}\Big]$
$=\frac{1-\cos^2\theta\sin^2\theta}{2+\cos^2\theta\sin^2\theta}$
$=\text{R.H.S}$
$\therefore\ \text{R.H.S.}=\text{L.H.S.}$