Question

Prove the identity $\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$.

Answer 1

Given,

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

$L.H.S=\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }$

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

$=\frac{\sin \theta }{\cos \theta }\left[\frac{1-2{\sin }^2\theta }{2(1-{\sin }^2\theta )-1}\right]$

$=\tan \theta \left[\frac{1-2{\sin }^2\theta }{2-2{\sin }^2\theta -1}\right]$

$=\tan \theta \left[\frac{1-2{\sin }^2\theta }{1-2{\sin }^2\theta }\right]$

$=\tan \theta$

$=R.H.S$

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