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

L.H.S= (1 ^2x- ^2x +1 cosec^2x- ^2x ) ^2x ^2x = (1 1 ^2x - ^2x +1 ^2x- ^2x ) ^2x ^2x (1 1- ^4x ^2x +1 1- ^4x ^2x ) ^2x ^2x = ( ^2x (1- ^2x)(1+ ^2x) + ^2x (1- ^2x)(1+ ^2x) ) ^2x ^2x ( c Using1-a^4=1-(a^2)^2 =(1-a^2)(1+a^2) ) = ( ^2x ^2x(1+ ^2x) + ^2x ^2x(1+ ^2) ) ^2x ^2x ( c Using 1- ^2x= ^2x &1 ^2x= ^2x ) = ( ^4x(1+ ^x)+ ^4x(1+ ^2x) ^2x ^2x(1+ ^2x)(1+ ^2x) ) ^2x ^2x = ^4x+ ^2x ^4x+ ^4x+ ^2x ^4x (1+ ^2x)(1+ ^2x) = ( ^2x)^2( ^2x)^2+2 ^2x ^2x-2 ^2x ^2x+ ^2x ^4x+ ^2x ^4x (1+ ^2x)(1+ ^2x) (adding an subtracting 2 ^2x ^4x) = ( ^2+ ^2x)^2-2 ^2x ^2x+ ^2x ^2x( ^2x+ ^2x) 1+ ^2x+ ^2x+ ^2x ^2x = 1^2-2 ^2x ^2x+ ^2x ^2x.1 1+1+ ^2x+ ^2x = 1- ^2x ^2x 2+ ^2x ^2x =R.H.S Proved

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

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Question

Prove the following identities: $\Big(\frac{1}{\sec^2\text{x}-\cos^\text{x}}+\frac{1}{\text{Cosec}^2\text{x}-\sin^2\text{x}}\Big)\sin^2\text{x}\cos^2\text{x}=\frac{1-\sin^2\text{x}\cos^2\text{x}}{2+\sin^2\text{x}\cos^2\text{x}}$

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Answer

$\text{L.H.S}=\Big(\frac{1}{\sec^2\text{x}-\cos^2\text{x}}+\frac{1}{\text{cosec}^2\text{x}-\sin^2\text{x}}\Big)\sin^2\text{x}\cos^2\text{x}$
$=\Bigg(\frac{1}{\frac{1}{\cos^2\text{x}}-\cos^2\text{x}}+\frac{1}{\sin^2\text{x}-\sin^2\text{x}}\Bigg)\sin^2\text{x}\cos^2\text{x}$
$\Bigg(\frac{1}{\frac{1-\cos^4\text{x}}{\cos^2\text{x}}}+\frac{1}{\frac{1-\sin^4\text{x}}{\sin^2\text{x}}}\Bigg)\sin^2\text{x}\cos^2\text{x}$
$=\Big(\frac{\cos^2\text{x}}{(1-\cos^2\text{x})(1+\cos^2\text{x})}+\frac{\sin^2\text{x}}{(1-\sin^2\text{x})(1+\sin^2\text{x})}\Big)\sin^2\text{x}\cos^2\text{x}$
$\left(\begin{array}{c}\text{Using}1-\text{a}^4=1-(\text{a}^2)^2\\ =(1-\text{a}^2)(1+\text{a}^2)\end{array}\right)$ $=\Big(\frac{\cos^2\text{x}}{\sin^2\text{x}(1+\cos^2\text{x})}+\frac{\sin^2\text{x}}{\cos^2\text{x}(1+\sin^2)}\Big)\sin^2\text{x}\cos^2\text{x}$
$\left(\begin{array}{c}\text{Using }1-\cos^2\text{x}=\sin^2\text{x}\\\&1 \sin^2\text{x}=\cos^2\text{x}\end{array}\right)$
$=\Big(\frac{\cos^4\text{x}(1+\sin^\text{x})+\sin^4\text{x}(1+\cos^2\text{x})}{\sin^2\text{x}\cos^2\text{x}(1+\cos^2\text{x})(1+\sin^2\text{x})}\Big)\sin^2\text{x}\cos^2\text{x}$ $=\frac{\cos^4\text{x}+\sin^2\text{x}\cos^4\text{x}+\sin^4\text{x}+\cos^2\text{x}\sin^4\text{x}}{(1+\cos^2\text{x})(1+\sin^2\text{x})}$
$=\frac{(\cos^2\text{x})^2(\sin^2\text{x})^2+2\cos^2\text{x}\sin^2\text{x}-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^4\text{x}+\cos^2\text{x}\sin^4\text{x}}{(1+\cos^2\text{x})(1+\sin^2\text{x})}$
$(\text{adding an subtracting }2\cos^2\text{x}\sin^4\text{x})$
$=\frac{(\cos^2+\sin^2\text{x})^2-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^2\text{x}(\cos^2\text{x}+\sin^2\text{x})}{1+\sin^2\text{x}+\cos^2\text{x}+\sin^2\text{x}\cos^2\text{x}}$
$=\frac{1^2-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^2\text{x}.1}{1+1+\sin^2\text{x}+\cos^2\text{x}}$ $=\frac{1-\sin^2\text{x}\cos^2\text{x}}{2+\sin^2\text{x}\cos^2\text{x}}$
$=\text{R.H.S} $ Proved