# Prove that (1-2sin^2x)/(1+sin2x) = (1-tanx)/(1+tanx)?

Apr 10, 2016

#### Explanation:

To prove $\frac{1 - 2 {\sin}^{2} x}{1 + \sin 2 x} = \frac{1 - \tan x}{1 + \tan x}$, let us start from left hand side.

$\frac{1 - 2 {\sin}^{2} x}{1 + \sin 2 x}$ is equivalent to

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

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

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

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

Now dividing numerator and denominator by $\cos x$

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

= $\frac{1 - \tan x}{1 + \tan x}$