Question : The expression $(\tan \theta+\cot \theta)(\sec \theta+\tan \theta)(1-\sin \theta), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\sec \theta$
Option 2: $\operatorname{cosec} \theta$
Option 3: $\cot \theta$
Option 4: $\sin \theta$
Correct Answer: $\operatorname{cosec} \theta$
Solution :
Using the trigonometric identities $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, and $1 - \sin^2 \theta = \cos^2 \theta$.
Substituting these identities into the expression,
$=(\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta})(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta})(1-\sin \theta)$
Simplifying this expression,
$=(\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta})(\frac{\sin \theta + 1}{\cos \theta})(1-\sin \theta)$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,
$=(\frac{1}{\sin \theta \cos \theta})(\frac{(1-\sin ^2\theta)}{\cos \theta})$
$=(\frac{1}{\sin \theta \cos \theta})(\frac{(cos ^2\theta)}{\cos \theta})$
$=\frac{1}{\sin \theta} $
$=\operatorname{cosec} \theta$
Hence, the correct answer is $\operatorname{cosec} \theta$.
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