Question

Let's see who can do this...

Answer 1

Step-by-step explanation:

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

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

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

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

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

$=\frac{sinA}{cosA}$

$=\boxed{tanA}$

Hope it helps!