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

Dec 31, 2015

Use identities and simplify. Try simplifying Left Hand Side (LHS) and Right Hand Side separately if you are unable to progress. It might end up giving same result.

#### Explanation:

Before trying to prove this review the following identities.

1) $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$
2) ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$
3) $1 - {\sin}^{2} \left(\theta\right) = {\cos}^{2} \left(\theta\right)$
4) $\cos \left(2 \theta\right) = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$
5) $\cos \left(2 \theta\right) = 2 {\cos}^{2} \left(\theta\right) - 1$
6)$\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \left(\theta\right)$
(7) 2sin^2(theta) = 1-cos(2theta))

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

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

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

LHS = RHS thus proved.