# How do I prove that sin^2x + tan^2xsin^2x = tan^2x ?

Sep 20, 2015

Use $\tan x = \sin \frac{x}{\cos} x$ and ${\sin}^{2} x + {\cos}^{2} x = 1$ and rearrange.

#### Explanation:

By definition:

$\tan x = \sin \frac{x}{\cos} x$

and by Pythagoras:

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

So:

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

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

$= {\sin}^{2} x + {\sin}^{2} x {\tan}^{2} x$

Sep 20, 2015

Using the rules: $\tan x = \sin \frac{x}{\cos} x$

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

#### Explanation:

${\sin}^{2} x + {\tan}^{2} x . {\sin}^{2} x = {\tan}^{2} x$

${\sin}^{2} x + {\sin}^{4} \frac{x}{{\cos}^{2} x} = {\tan}^{2} x$

${\sin}^{2} x + {\sin}^{4} \frac{x}{{\cos}^{2} x} = {\sin}^{2} \frac{x}{{\cos}^{2} x}$

${\sin}^{2} x = \frac{{\sin}^{2} x - {\sin}^{4} x}{{\cos}^{2} x}$

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

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

${\sin}^{2} x = {\sin}^{2} x$