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

Nov 4, 2016

Verifying the equation of trigonometric identities is determined by applying some properties

$\textcolor{red}{\tan x = \sin \frac{x}{\cos} x}$

$\textcolor{b l u e}{\tan \left(- x\right) = - \tan x}$

$\textcolor{p u r p \le}{{\cos}^{2} x + {\sin}^{2} x = 1}$

Verifying the equality:
${\tan}^{2} \left(- x\right) {\cos}^{2} x = {\left(\tan \left(- x\right)\right)}^{2} {\cos}^{2} x$

${\tan}^{2} \left(- x\right) {\cos}^{2} x = {\left(\textcolor{b l u e}{- \tan \left(x\right)}\right)}^{2} {\cos}^{2} x$

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

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

tan^2(-x)cos^2x=color(red)((sinx/cosx)^2cos^2x

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

tan^2(-x)cos^2x=color(violet)(sin^2x

We have
$\textcolor{p u r p \le}{{\cos}^{2} x + {\sin}^{2} x = 1}$
rArrcolor(violet)(sin^2x=1-cos^2x

Therefore,
tan^2(-x)cos^2x=color(violet)(1-cos^2x