# How do you verify the identity (tanx-secx)/(tanx+secx)=(tan^2x-1)/(sec^2x)?

Apr 5, 2018

We have:

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

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

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

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

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

$- 2 \sin x - {\cos}^{2} x = {\sin}^{2} x - {\cos}^{2} x$

Since $- 2 \sin x \ne {\sin}^{2} x$, this identity is false.

Hopefully this helps!