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

Mar 28, 2018

#### Explanation:

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

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

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

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

$= \frac{1}{1 - {\tan}^{2} x} - {\cos}^{2} x$,

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

$= \left\{1 \div \frac{{\cos}^{2} x - {\sin}^{2} x}{\cos} ^ 2 x\right\} - {\cos}^{2} x$,

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

Mar 28, 2018

${\sin}^{2} x + \frac{\textcolor{red}{2} {\tan}^{2} x}{1 - {\tan}^{2} x} = \frac{1}{{\cos}^{2} x - {\sin}^{2} x} - {\cos}^{2} x$ ?

#### Explanation:

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

$L H S = {\sin}^{2} x + \frac{2 {\tan}^{2} x}{1 - {\tan}^{2} x}$

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

$= \left[1 + \frac{2 {\tan}^{2} x}{1 - {\tan}^{2} x}\right] - {\cos}^{2} x$

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

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

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

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

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

$= R H S$