Question : If $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{1}{7}, \theta$ lies in first quadrant, then the value of $\frac{\operatorname{cosec} \theta+\cot ^2 \theta}{\operatorname{cosec} \theta-\cot ^2 \theta}$ is:
Option 1: $\frac{19}{5}$
Option 2: $\frac{22}{3}$
Option 3: $\frac{37}{12}$
Option 4: $\frac{37}{19}$
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Correct Answer: $\frac{19}{5}$
Solution :
$\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{1}{7}$
Using the identity $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$,
$\frac{\frac{1}{cos\theta}-\frac{\sin \theta}{\cos \theta}}{\frac{1}{cos\theta}+\frac{\sin\theta}{\cos\theta}}=\frac{1}{7}$
$⇒\frac{1-\sin \theta}{1+\sin \theta}=\frac{1}{7}$
On applying componendo and dividendo,
$⇒\sin \theta = \frac{3}{4}$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$,
$⇒\cos \theta = \frac{\sqrt{7}}{4}$
Substituting these values into the expression $\frac{\operatorname{cosec} \theta+\cot ^2 \theta}{\operatorname{cosec} \theta-\cot ^2 \theta}$,
$=\frac{\frac{4}{3}+\left(\frac{\sqrt{7}}{3}\right)^2}{\frac{4}{3}-\left(\frac{\sqrt{7}}{3}\right)^2} =\frac{\frac{4}{3}+\frac{7}{9}}{\frac{4}{3}-\frac{7}{9}}
= \frac{19}{5}$
Hence, the correct answer is $\frac{19}{5}$.
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