Question

Prove that

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

Answer 1

Here, $\frac{2\sin \theta -\sin 2\theta }{2\sin \theta +\sin 2\theta }=\frac{2\sin \theta -2\sin \theta \cos \theta }{2\sin \theta +2\sin \theta \cos \theta }$ $[\because \sin 2\theta =2\sin \theta \cos \theta ]$

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

$=\frac{1-\cos \theta }{1+\cos \theta }$

$=\frac{2{\sin }^2\theta /2}{2{\cos }^2\theta /2}$ $\because \left[\begin{array}{c}\cos 2\theta =1-2{\sin }^2\theta \\ \cos 2\theta =2-{\cos }^2\theta -1\end{array}\right]$

$={\tan }^2\theta /2$

$=1-\frac{2\tan \theta /2}{\tan \theta }$ $\left[\begin{array}{c}\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\\ \Rightarrow 1-{\tan }^2\theta =\frac{2\tan \theta }{{\tan }^2\theta }\\ \Rightarrow {\tan }^2\theta =1-\frac{2\tan \theta }{{\tan }^2\theta }\end{array}\right]$.