# How do you prove  tan^2x(1-sin^2x)=(-cos2x+1)/2 ?

$L H S$

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

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

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

$= 1 - \setminus {\cos}^{2} x$

$1 - \left(\setminus \frac{1 + \setminus \cos 2 x}{2}\right)$

$= \setminus \frac{2 - 1 - \setminus \cos 2 x}{2}$

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

$= R H S$

Proved.

Jul 1, 2018

#### Explanation:

We know that,

color(red)((1)sin^2theta+cos^2theta=1

color(blue)((2)1-cos2theta=2sin^2theta

We take ,

LHS=tan^2x(color(red)(1-sin^2x))tocolor(red)(Apply(1)

$\textcolor{w h i t e}{L H S} = {\sin}^{2} \frac{x}{\cos} ^ 2 x \left(\textcolor{red}{{\cos}^{2} x}\right) \to \left[\because \tan \theta = \sin \frac{\theta}{\cos} \theta\right]$

$\textcolor{w h i t e}{L H S} = {\sin}^{2} x$

color(white)(LHS)=1/2[color(blue)(2sin^2x)]tocolor(blue)(Apply(2)

$\textcolor{w h i t e}{L H S} = \frac{1}{2} \left[\textcolor{b l u e}{1 - \cos 2 x}\right]$

$\textcolor{w h i t e}{L H S} = \frac{- \cos 2 x + 1}{2}$

$\implies L H S = R H S$