Question : The expression $\frac{\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)(\cot \theta+1) \cos \theta}{\left(\sin ^4 \theta+\cos ^4 \theta\right)(1+\tan \theta) \operatorname{cosec} \theta}-1,0^{\circ}<\theta<90^{\circ}$, equals:
Option 1: $\cos ^2 \theta$
Option 2: $-\sin ^2 \theta$
Option 3: $\sec ^2 \theta$
Option 4: $-\sec ^2 \theta$
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Correct Answer: $-\sin ^2 \theta$
Solution :
Given: $\frac{\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)(\cot \theta+1) \cos \theta}{\left(\sin ^4 \theta+\cos ^4 \theta\right)(1+\tan \theta) \operatorname{cosec} \theta}-1,0^{\circ}<\theta<90^{\circ}$
$=\frac{\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)(\frac{\cos \theta}{\sin \theta}+1) \cos \theta}{\left((\sin ^2 \theta+\cos ^2 \theta)^2-2 \sin ^2 \theta \cos ^2 \theta\right)(\frac {\sin \theta}{\cos \theta}+1) \operatorname{cosec} \theta}-1$
$=\frac{\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)(\frac{\cos \theta+\sin \theta}{\sin \theta}) \cos \theta}{\left(1-2 \sin ^2 \theta \cos ^2 \theta\right)(\frac {\sin \theta+\cos \theta}{\cos \theta}) \operatorname{cosec} \theta}-1$
$=\frac{(\frac{\cos \theta+\sin \theta}{\sin \theta}) \cos \theta}{(\frac {\sin \theta+\cos \theta}{\cos \theta}) \operatorname{cosec} \theta}-1$
$=\cos ^2\theta-1$
$=-\sin ^2\theta$
Hence, the correct answer is $-\sin ^2\theta$.
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