$\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$ is equal to
(a) 2 tan θ
(b) 2 sec θ
(c) 2 cosec θ
(d) 2 tan θ sec θ
The given expression is $\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$.
Simplifying the given expression, we have
$\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}$
$=\frac{\tan \theta(\sec \theta+1)+\tan \theta(\sec \theta-1)}{(\sec \theta-1)(\sec \theta+1)}$
$=\frac{\tan \theta \sec \theta+\tan \theta+\tan \theta \sec \theta-\tan \theta}{\sec ^{2} \theta-1}$
$=\frac{2 \tan \theta \sec \theta}{\tan ^{2} \theta}$
$=\frac{2 \sec \theta}{\tan \theta}$
$=\frac{2 \frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}$
$=2 \frac{1}{\sin \theta}$
$=2 \operatorname{cosec} \theta$
Therefore, the correct option is (c).
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