Question

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

Answer 1

Simplify left hand side of $\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\tan \theta$

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

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

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

$=\tan \theta$

$=RHS$