Question

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

Answer 1

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

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

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

$=$R.H.S