Question

$\frac{\tan \theta }{\sec \theta -1}+\frac{\tan \theta }{\sec \theta +1}$ is equal to

• A. $\text{cosec}\theta$
• B. $2\text{cosec}\theta$
• C. $\sec \theta$
• D. $2\sec \theta$

Answer 1

Answer: (B)

$2\text{cosec}\theta$

$\frac{\tan \theta (\sec \theta +1+\sec \theta -1)}{{\sec }^2\theta -1}=\frac{2\tan \theta \sec \theta }{{\sec }^2\theta -1}=\frac{2\tan \theta \sec \theta }{{\tan }^2\theta }$
$\Rightarrow \frac{2\sec \theta }{\tan \theta }=2\frac{1}{\cos \theta .\frac{\sin \theta }{\cos \theta }}=2\text{cosec}\theta$