Question

$\frac{2\tan \theta (1-{\tan }^2\theta )}{\sec \theta {(1+\tan \theta )}^2}$ equal to

• A. $\frac{2\sin \theta (1-\tan \theta )}{(1+\tan \theta )}$
• B. $\frac{2\sin \theta }{1+\tan \theta }$
• C. $\frac{2\sin \theta (1+\tan \theta )}{(1-\tan \theta )}$
• D. None of these

Answer 1

The correct option is A $\frac{2\sin \theta (1-\tan \theta )}{(1+\tan \theta )}$

Consider the given expression.

$\frac{2\tan \theta (1-{\tan }^2\theta )}{\sec \theta {(1+\tan \theta )}^2}$

$\Rightarrow \frac{\frac{2\sin \theta }{\cos \theta }(1-{\tan }^2\theta )}{\frac{1}{\cos \theta }{(1+\tan \theta )}^2}[\because {\sec }^2\theta =1+{\tan }^2\theta ]$

$\Rightarrow \frac{2\sin \theta (1-\tan \theta )(1+\tan \theta )}{{(1+\tan \theta )}^2}$

$\Rightarrow \frac{2\sin \theta (1-\tan \theta )}{(1+\tan \theta )}$

Hence, this is the answer.