How do you prove cosx- (cosx/(1-tanx))= (sinxcosx)/(sinx-cosx)?

Feb 17, 2016

Please see below for the proof. Feel free to ask questions if you have any.

Explanation:

1) replace tanx with $\sin \frac{x}{\cos} x$

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

2) equalise the denominator in the paranthesis

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

=$\cos x - \left(\cos x \cdot \cos \frac{x}{\cos x - \sin x}\right)$
=$\cos x - {\cos}^{2} \frac{x}{\cos x - \sin x}$

3) equalise the denominator once more
=$\cos x \frac{\cos x - \sin x}{\cos x - \sin x} - {\cos}^{2} \frac{x}{\cos x - \sin x}$
=$\frac{{\cos}^{2} x - \cos x \cdot \sin x - {\cos}^{2} x}{\cos x - \sin x}$
=$\frac{- \left(\cos x \cdot \sin x\right)}{\cos x - \sin x}$

4) put the denominator into -1 paranthesis
=$- \frac{\cos x \cdot \sin x}{- \left(- \cos x + \sin x\right)}$

5) - divided by - yields +. And change the places of cosx and sinx in the denominator

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

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