-1 + + 1 is equal to :

MCQ

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

A
$2\tan\theta$
B
$2\sec\theta$
✓
$2\text{cosec }\theta$
D
$2\tan\theta\sec\theta$

✓

Answer

Correct option: C.

$2\text{cosec }\theta$

The givne expression is $\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
Simplifying the given expression, we have
$\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
$=\frac{\tan\theta(\sec\theta+1)+\tan\theta(\sec\theta-1)}{(\sec\theta-1)(\sec\theta+1)}$
$=\frac{\tan\theta\sec\theta+\tan\theta+\tan\theta\sec\theta-\tan\theta}{\sec^2\theta-1}$
$=\frac{2\tan\theta\sec\theta}{\tan^2\theta}$
$=\frac{2\sec\theta}{\tan\theta}$
$=\frac{2\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=2\frac{1}{\sin\theta}$
$=2\text{cosec }\theta$
Therefore, the correct option is $(c).$

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