can anyone please help me prove the following:

sin^3x cos^3x/1-2cos^2x = secx-sinx/tanx-1

1-2\cos^2 x=\sin^2 x+\cos^2 x-2\cos^2 x=\sin^2 x-\cos^2 x

By direct computation: computing the left hand side you have (using 1-2\cos^2x=\sin^2x+\cos^2x-2\cos^2x )

\frac{1-\sin x\cos x}{\sin x-\cos x}

Right hand side is

\frac{\cos^{-1}x-\sin x }{\tan x-1}=\frac{1-\cos x\sin x}{\cos x(\tan x-1)}

Then you conclude by using \sin x=\cos x\tan x .