Question

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

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

Answer 1

Answer: (B)

$\frac{2\sin \theta }{1+\tan \theta }$

Consider the given expression.

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

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

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

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

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

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

Hence, this is the answer.