Question

Show that:

$\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$

Answer 1

$\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$

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

$\Rightarrow \frac{1-2{\sin }^2\theta }{2{\cos }^2\theta -1}=1$

$\Rightarrow 1-2{\sin }^2\theta =2{\cos }^2\theta -1$

$\Rightarrow \cos 2\theta =\cos 2\theta (\because \cos 2\theta =1-2{\sin }^2\theta =2{\cos }^2\theta -1)$

L.H.S. $=$ R.H.S.

Hence proved.