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

Feb 13, 2016

The identity is false.

#### Explanation:

The given identity is not true. For example, if $x = \frac{\pi}{6}$

$\frac{\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{6}\right)}{{\cos}^{2} \left(\frac{\pi}{6}\right) - {\sin}^{2} \left(\frac{\pi}{6}\right)} = \frac{\frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\frac{3}{4} - \frac{1}{4}} = \frac{\frac{\sqrt{3}}{4}}{\frac{2}{4}} = \frac{\sqrt{3}}{2}$

but

$\tan \frac{\frac{\pi}{6}}{1 - \tan \left(\frac{\pi}{6}\right)} = \frac{\frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{1}{\sqrt{3} - 1} \ne \frac{\sqrt{3}}{2}$

In fact, the equation is only true when $\sin \left(x\right) = 0$ as can be seen here:

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

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

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

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

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

thus the equation holds only when ${\sin}^{2} \frac{x}{{\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)} = 0$
which is true only when $\sin \left(x\right) = 0$