Question

pls.....slove it...step by step ​

Answer 1

Step-by-step explanation:

For the better understanding, I am replacing θ by A.

LHS:

$Given:\frac{sinA-2sin^3A}{2cos^3A-cosA}$

$=\frac{sinA(1-2sin^2A)}{cosA(2cos^2A-1)}$

$=tanA * \frac{1-2sin^2A}{2cos^2A-1}$

$=tanA*\frac{1-2sin^2A}{2(1-2sin^2A)-1}$

$=tanA*\frac{1-2sin^2A}{2 - 2sin^2A - 1}$

$=tanA * \frac{1-2sin^2A}{1-2sin^2A}$

$=\boxed{tanA}$

RHS

Hope it helps!