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:
Option 1: $ \sin \theta+\cos \theta$
Option 2: $\sin \theta \cos \theta$
Option 3: $\sec \theta \operatorname{cosec} \theta$
Option 4: $\sec \theta+\operatorname{cosec} \theta$
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Correct Answer: $\sec \theta \operatorname{cosec} \theta$
Solution : $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ $=\operatorname{cosec}^2\theta×\sec^2\theta(\sin\theta-\frac{1}{\sin\theta})(\cos\theta-\frac{1}{\cos\theta})$ [$\because 1+\cot ^2 \theta=\operatorname{cosec}^2\theta$ and $1+\tan ^2 \theta=\sec^2\theta$] $=\frac{1}{\sin^2\theta}×\frac{1}{\cos^2\theta}(\frac{\sin^2\theta-1}{\sin\theta})(\frac{\cos^2\theta-1}{\cos\theta})$ $=\frac{1}{\sin^2\theta}×\frac{1}{\cos^2\theta}(\frac{-\cos^2\theta}{\sin\theta})(\frac{-\sin^2\theta}{\cos\theta})$ $=\sec \theta \operatorname{cosec} \theta$ Hence, the correct answer is $\sec \theta \operatorname{cosec} \theta$.
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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 : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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:
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