Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

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

Answer 1

Proof:

Consider the LHS of the given expression.

$$\begin{gathered} \mathrm{LHS}=\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta } \\ =\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)} \\ =\frac{\sin \theta }{\cos \theta }·\left(\frac{1-2{\sin }^2\theta }{2(1-{\sin }^2\theta )-1}\right)\left[\mathrm{Use}{\cos }^2\theta =1-{\sin }^2\theta \right] \\ =\frac{\sin \theta }{\cos \theta }·\left(\frac{1-2{\sin }^2\theta }{2-2{\sin }^2\theta -1}\right) \\ =\frac{\sin \theta }{\cos \theta }·\left(\frac{1-2{\sin }^2\theta }{1-2{\sin }^2\theta }\right) \\ =\frac{\sin \theta }{\cos \theta } \\ =\tan \theta \left[\because \frac{\sin \theta }{\cos \theta }=\tan \theta \right] \\ =\mathrm{RHS} \end{gathered}$$

Hence, proved.