# How do you verify sin^2xsec^2x+sin^2xcsc^2x=sec^2x?

May 20, 2018

See below for verification with the assumption that $\sin \left(x\right) \ne 0$
The given equation is not true if $\sin \left(x\right) = 0$

#### Explanation:

${\sin}^{2} \left(x\right) {\sec}^{2} \left(x\right) + {\sin}^{2} \left(x\right) {\csc}^{2} \left(x\right)$

$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{2} \left(x\right) \left[{\sec}^{2} \left(x\right) + {\csc}^{2} \left(x\right)\right]$

$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{2} \left(x\right) \left[\frac{1}{\cos} ^ 2 \left(x\right) + \frac{1}{{\sin}^{2} \left(x\right)}\right]$

$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{2} \left(x\right) \left[\frac{{\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)}{{\cos}^{2} \left(x\right) {\sin}^{2} \left(x\right)}\right]$

$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{2} \left(x\right) \left[\frac{1}{{\cos}^{2} \left(x\right) {\sin}^{2} \left(x\right)}\right]$

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{{\cos}^{2} \left(x\right)}$

$\textcolor{w h i t e}{\text{XXX}} = {\sec}^{2} \left(x\right)$