# How do you prove cos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)?

Aug 12, 2015

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

#### Explanation:

Assume that $\frac{\cos 2 x}{1 - \sin 2 x} = \frac{1 + \tan 2 x}{1 - \tan 2 x}$ is true for all $x$

Taking $2 x = \frac{\pi}{4}$

LHS: $\cos \frac{\frac{\pi}{4}}{1 - \sin \left(\frac{\pi}{4}\right)} = \frac{\sqrt{2}}{2 - \sqrt{2}}$
RHS: ${\lim}_{2 x \rightarrow \frac{\pi}{4}} \frac{1 + \tan \left(2 x\right)}{1 - \tan \left(2 x\right)} = \infty$

Therefore $\frac{\cos 2 x}{1 - \sin 2 x} \ne \frac{1 + \tan 2 x}{1 - \tan 2 x}$