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K_T.T. · Nov 20, 2011, 04:05 PM

can anyone please help me prove the following:

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

Aurora2000 · Nov 21, 2011, 03:29 AM

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

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 $\sin x=\cos x\tan x$ .