# How do you show that cosx/(1 - tanx) - sin^2x/(cosx- sinx) = cosx + sinx?

Oct 31, 2016

The following identities are important for this problem:

-$\tan \theta = \sin \frac{\theta}{\cos} \theta$

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

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

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

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

Factor the numerator on the left as a difference of squares, ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$.

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

$\cos x + \sin x = \cos x + \sin x$

Identity Proved!!

Hopefully this helps!

Oct 31, 2016

$L H S = \cos \frac{x}{1 - \tan x} - {\sin}^{2} \frac{x}{\cos x - \sin x}$

$= {\cos}^{2} \frac{x}{\cos x \left(1 - \sin \frac{x}{\cos} x\right)} - {\sin}^{2} \frac{x}{\cos x - \sin x}$

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

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

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

$= \cos x + \sin x = R H S$

Proved