Question

$\frac{[2\sin \theta \tan \theta (1-\tan \theta )+2\sin \theta se{c}^2\theta ]}{{(1+\tan \theta )}^2}$

• 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. $Noneofthese$

Answer 1

The correct option is B

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

Explanation for the correct option:

Solving the given expression:

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

$=\frac{2\sin \theta \tan \theta -2\sin \theta {\tan }^2\theta +2\sin \theta (1+{\tan }^2\theta )}{{(1+\tan \theta )}^2}$ $\left[\mathbf{\because }\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{x}\right]$

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

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

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

$=\frac{\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{}}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }\mathbf{)}}$

Hence, Option ‘B’ is Correct.