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:
Option 1: $\operatorname{cosec} \theta \sec \theta$
Option 2: $\operatorname{cosec} \theta$
Option 3: $\sin \theta \cos \theta$
Option 4: $\sec \theta$
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Correct Answer: $\sin \theta \cos \theta$
Solution :
$\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right)$
= $(\frac{ \frac{\sin^3 \theta}{\cos^3 \theta}}{\frac{1}{\cos^2 \theta}}+\frac{\frac{\cos^3 \theta}{\sin^3 \theta}}{\frac{1}{\sin^2 \theta} \theta}+2 \sin \theta \cos \theta) \div(1+\tan ^2 \theta+\operatorname{cosec}^2 \theta)$
= $\left(\frac{\sin^3 \theta}{\cos\theta} + \frac{\cos^3 \theta}{\sin\theta} + 2 \sin \theta \cos \theta\right) \div (\sec^2 \theta+\operatorname{cosec}^2 \theta)$
= $(\frac{\sin^4 \theta + \cos^4 \theta \sin +2 \sin^2 \theta \cos^2 \theta}{\sin\theta\cos\theta}) \div (\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta})$
= $\frac{(\sin^2\theta+\cos^2\theta)^2}{\sin\theta\cos\theta}×\frac{\sin^2\theta\cos^2\theta}{\sin^2\theta+\cos^2\theta}$
= $\sin\theta\cos\theta$
Hence, the correct answer is $\sin\theta\cos\theta$.
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