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

Oct 27, 2016

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

$\frac{{\tan}^{2} x}{{\tan}^{2} x + 1} = {\sin}^{2} x$

$\Rightarrow \frac{{\tan}^{2} x}{{\tan}^{2} x + 1} = {\sin}^{2} \left(- x\right)$.

#### Explanation:

Verifying the equality of the given trigonometric expressions is determined by evaluating each side separately then comparing their results.

These trigonometric identities are used
$\textcolor{red}{\sin \left(- x\right) = - \sin x}$

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

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

${\sin}^{2} \left(- x\right) = {\left(\sin \left(- x\right)\right)}^{2} = {\left(\textcolor{red}{- \sin x}\right)}^{2} = {\sin}^{2} x$
$\textcolor{b l u e}{{\sin}^{2} \left(- x\right) = {\sin}^{2} x \text{ } E Q 1}$

$\frac{{\tan}^{2} x}{{\tan}^{2} x + 1}$

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

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

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

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

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x \cdot {\cos}^{2} \frac{x}{1}$

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

$\textcolor{b l u e}{\frac{{\tan}^{2} x}{{\tan}^{2} x + 1} = {\sin}^{2} x}$

$\textcolor{b l u e}{{\sin}^{2} \left(- x\right) = {\sin}^{2} x \text{ } E Q 1}$

Therefore,

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