Question

Prove that:

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

Answer 1

L.H.S.

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

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

We know that

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

Therefore,

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

$=\frac{\sin \theta }{\cos \theta }$

$=\tan \theta$

R.H.S

Hence, proved.