# How can you show tan^2 x(1+cos2x)=2sin^2x?

We can use the $\textcolor{b l u e}{\text{double angle identity for cos(2x)}}$ as well as the $\textcolor{g r e e n}{\text{ratio identity for tan x}}$.
${\tan}^{2} x \left(1 + \textcolor{b l u e}{\cos 2 x}\right) = {\tan}^{2} x \left(\cancel{1} + \textcolor{b l u e}{2 {\cos}^{2} x - \cancel{1}}\right)$
$\textcolor{w h i t e}{{\tan}^{2} x \left(1 + \cos 2 x\right)} = {\tan}^{2} x \left(2 {\cos}^{2} x\right)$
$\textcolor{w h i t e}{{\tan}^{2} x \left(1 + \cos 2 x\right)} = 2 \textcolor{g r e e n}{{\tan}^{2} x} {\cos}^{2} x$
$\textcolor{w h i t e}{{\tan}^{2} x \left(1 + \cos 2 x\right)} = 2 \textcolor{g r e e n}{\left({\sin}^{2} \frac{x}{\cancel{{\cos}^{2} x}}\right)} \cancel{{\cos}^{2} x}$
$\textcolor{w h i t e}{{\tan}^{2} x \left(1 + \cos 2 x\right)} = 2 {\sin}^{2} x$