I have the following trigonometric identity

$$\cos{x} - \frac{\cos{x}}{1 - \tan{x}} = \frac{\sin{x} \cos{x}}{\sin{x} - \cos{x}}$$

I've been trying to verify it for almost 20 minutes but coming up with nothing

Thank you


Observe that we need to eliminate $\tan x$

So, using $\tan x=\frac{\sin x}{\cos x},$

$$\cos x-\frac{\cos x}{1-\tan x}$$

$$=\cos x\left(1-\frac{1}{1-\frac{\sin x}{\cos x}}\right)$$

$$=\cos x\left(1-\frac{\cos x}{\cos x-\sin x}\right) (\text{ multiplying numerator & denominator by }\cos x)$$

$$=\cos x\left(1+\frac{\cos x}{\sin x-\cos x}\right)$$

$$=\cos x\left(\frac{\sin x}{\sin x-\cos x}\right)$$

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  • $\begingroup$ How is your final answer equal to the RHS? $\endgroup$ – user82124 Jun 12 '13 at 15:28
  • $\begingroup$ @user82124, how about multiplying $\cos x$ and $\left(\frac{\sin x}{\sin x-\cos x}\right)$ $\endgroup$ – lab bhattacharjee Jun 12 '13 at 15:29
  • $\begingroup$ That would give cos x sin x / sin x cos x^2 $\endgroup$ – user82124 Jun 12 '13 at 15:31
  • $\begingroup$ @user82124, where is this $\cos x\sin x/\sin x\cos x^2$ coming from? $\endgroup$ – lab bhattacharjee Jun 12 '13 at 15:32
  • $\begingroup$ From multiplying cos x and ( Sin x / Sin x - Cos X )? $\endgroup$ – user82124 Jun 12 '13 at 15:35

$$ \frac{\cos}{1-\tan x}\cdot\frac{\cos x}{\cos x} = \frac{\cos^2 x}{\cos x-\sin x} $$

$$ \cos x\cdot\frac{\cos x-\sin x}{\cos x-\sin x} - \frac{\cos^2 x}{\cos x-\sin x} = \frac{-\sin x\cos x}{\cos x-\sin x} $$

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