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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Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : Let $0^{\circ}<\theta<90^{\circ}$, $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ is equal to:
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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