Question

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

Answer 1

$\begin{array}{cc}\frac{2\sin \theta \cdot \tan \theta \left(1-\tan \theta \right)+2\sin \theta \cdot {\sec }^2\theta }{{\left(1+\tan \theta \right)}^2}& \\ \Rightarrow \frac{2\sin \theta \left[\tan \theta -{\tan }^2\theta +1\times {\sec }^2\theta \right]}{{\left(1+\tan \theta \right)}^2}& \\ \Rightarrow \frac{2\sin \theta \left[\tan \theta -{\tan }^2\theta +{\sec }^2\theta \right]}{{\left(1+\tan \theta \right)}^2}& \left[{\sec }^2\theta \cdot {\tan }^2\theta =1\right]\\ \Rightarrow \frac{2\sin \theta }{{\left(1+\tan \theta \right)}^2}\left[1+\tan \theta \right]& \\ \therefore \frac{2\sin \theta }{\left(1+\tan \theta \right)}& \end{array}$