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