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

Given :

The given expression is $\frac{sin \theta-2 sin ^{3} \theta}{2 cos ^{3} \theta-cos \theta}= tan \theta$.

To do :

We have to prove the LHS and RHS are equal in the given expression.

Solution :

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)} $ [Taking $sin \theta$ as common in the numerator and $cos \theta$ as common in the denominator]

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

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

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

$ = tan \theta$

RHS $= tan \theta$

LHS $=$ RHS.

Hence proved.