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