Question

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

Answer 1

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

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

$=\frac{\sin \theta .\cos \theta }{\cos \theta .\cos 2\theta }$

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

$=\tan \theta$.