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

Apr 14, 2015

$\frac{\sin x \cos x}{\sin x - \cos x} = \cos x - \left[\frac{\cos x}{1 - \tan x}\right]$

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

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

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

$\frac{\sin x \cos x}{\sin x - \cos x} = \cos x - \left[\frac{{\cos}^{2} x}{\cos x - \sin x}\right]$

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

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

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

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

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

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