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

Mar 7, 2016

Starting from the right hand side:

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

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

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

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

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

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