Question

sinA -2sin³A/2cos³A-cosA=tanA

Answer 1

that 's so simple.
Please select as brainlist

Answer 2

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!