Question

$\frac{\cot \theta }{1+\tan \theta }=\frac{\cot \theta -1}{2-{\sec }^2\theta }$

• A. True
• B. False

Answer 1

The correct option is A True

LHS :

$\frac{\cot \theta }{1+\tan \theta }$

$=\frac{\cot \theta }{1+\tan \theta }\times \frac{1-\tan \theta }{1-\tan \theta }$

$=\frac{\cot \theta -\cot \theta \tan \theta }{1-{\tan }^2\theta }$

$=\frac{\cot \theta -1}{1-({\sec }^2\theta -1)}$

$=\frac{\cot \theta -1}{2-{\sec }^2\theta }$ = RHS