0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\tan^2(-x)=(\tan(-x))^2=(-\tan x)^2=\tan^2x$ $\endgroup$ – CY Aries Apr 29 '17 at 19:24
1
$\begingroup$

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 $-\sin^2-1$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $-1.$

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.