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:

Option 1: $\sin \theta$

Option 2: $2 \cos \theta$

Option 3: $\cot \theta$

Option 4: $2 \tan \theta$

Correct Answer: $2 \tan \theta$

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