# How to simplify a trigonometric expression with the identities

Simplify the trigonometric expression

$$\frac{\cos^{2}x\tan^{2}(-x)-1}{\cos^{2}x}.$$

For this math problem, I attempted to multiply $\cos^2 x$ and $\tan^2(-x)$ so it becomes $-\sin^2 (x) - 1$, but I am still stuck on how to simplify this problem.

Edit: Thanks to the commenter for helping me solve, but I just have one more question. Would the solution for this be -1? The numerator simplified to sin^2 x - 1, which is equivalent as -cos^2 x from the pythagorean identity. I'm just curious if -1 is the solution.

• $\tan^2(-x)=(\tan(-x))^2=(-\tan x)^2=\tan^2x$ – CY Aries Apr 29 '17 at 19:24

Note that $\tan (-x)=-\tan x$, so $\tan^2(-x)=\tan^2(x)$. Then $\cos x \tan x = \sin x$, so you dropped a sign when you say the numerator becomes $-\sin^2-1$

Using that $\tan^2(-x)=\tan^2(x)$ and that $1=\cos^2(x)+\sin^2(x)$, we get :$$\frac{\cos^{2}x\tan^{2}(-x)-1}{\cos^{2}x}=\frac{\cos^{2}x\tan^{2}(x)-\cos^2(x)-\sin^2(x)}{\cos^{2}x}\\ =\tan^2(x)-1-\tan^2(x)=-1$$

First, since $\sin (-x)=-\sin x$ and $\cos(-x)=\cos x,$ we have $\tan(-x)=\sin(-x)/\cos(-x)=(-\sin x)/\cos x=-\tan x.$

So $\tan^2(-x)=(\tan (-x))^2=(-\tan x)^2=(\tan x)^2=\tan^2 x.$

Second, $\cos^2x\tan^2(-x)=\cos^2x\tan^2x=(\cos^2x)(\sin x/\cos x)^2=(\cos x)^2((\sin x)^2/(\cos x)^2)=$ $=(\sin x)^2=\sin^2 x.$

So $\cos^2 x \tan^2(-x)-1=\sin^2x - 1.$

Third, $\sin^2x +\cos^2 x=1$ so $\sin^2x=1-\cos^2x$. So $\sin^2x-1=-\cos^2 x.$

Fourth, since the numerator in the expression is seen to be equal to $-\cos^2x$ and the denominator is $+\cos^2 x,$ the expression is equal to $-1.$

$$\frac{\cos^{2}x\tan^{2}(-x)-1}{\cos^{2}x}=\frac{\cos^{2}x\tan^{2}(+x)-1}{\cos^{2}x}\\ =\frac{\sin^{2}x-1}{\cos^{2}x}=-1.$$