# How do you prove (1 - tanx) / (1 + tanx) = (1 - sin2x) / (cos2x)?

Apr 27, 2015

Multiply the left side in the numerator and denominator by 1-tanx and simplify to get the answer:

${\left(1 - \tan x\right)}^{2} / \left(1 - {\tan}^{2} x\right)$

=(1+tan^2x -2tanx)/(1- (sin^2x/cos^2x)

= $\frac{{\sec}^{2} x - 2 \tan x}{\frac{{\cos}^{2} x - {\sin}^{2} x}{\cos} ^ 2 x}$ (because $1 + {\tan}^{2} x = {\sec}^{2} x$)

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

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