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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