Proving the identity $(\tan^2(x)+1)(\cos^2(-x)-1)=-\tan^2(x)$ Proving the trigonometric identity $(\tan{^2x}+1)(\cos{^2(-x)}-1)=-\tan{^2x}$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.
 A: Hint:
Multiply the two members by $\cos^2x$ (certainly nonzero):
$$(\sin^2x+\cos^2x)(\cos^2x-1)=-\sin^2x.$$
A: Expanding gives $\sin^2x-\tan^2x+\cos^2x-1$. The undesired terms cancel because $\sin^2x+\cos^2x=1$.
A: Recall:
$$\begin{align}
\tan^2(x) - \sec^2(x) &= -1 \\
\sin^2(x) + \cos^2(x) &= 1
\end{align}$$
Thus,
$$\begin{align}
\tan^2(x) + 1 &= \sec^2(x)\\
\cos^2(x) - 1 &= -\sin^2(x)
\end{align}$$
We also note that $\cos(x)$ is an even function, and thus $\cos(-x) = \cos(x)$. Thus, the formula becomes:
$$(\tan^2(x) + 1)(\cos^2(-x) - 1) = -\sec^2(x)\sin^2(x) = - \frac{\sin^2(x)}{\cos^2(x)} = -\tan^2(x)$$
A: *

*$1+\tan^2x=\sec^2x$

*$\sin^2x+\cos^2x=1$

*$\cos x=\cos(-x)$ i.e. $\cos x$ is an even function.

*$\sec x=1/\cos x$
$$\underbrace{\left(1+\tan^2x\right)}_{=\sec^2x}\underbrace{\left(\cos^2(-x)-1\right)}_{\cos x\text{ is even function}}=-\sec^2x\cdot\sin^2x=-\tan^2x $$
A: We know


*

*$(\tan x)' = 1 + \tan^2 x = \frac{1}{\cos^2 x}$ and

*$\cos (-x) = \cos x$
Now, it follows immediately
\begin{eqnarray*} (\tan^2(x)+1)(\cos^2(-x)-1)
& = & (\tan x)'\left(\frac{1}{(\tan x)'}-1\right) \\
& = & 1 - (\tan x)' \\
& = & 1 - (1 + \tan ^2 x) \\
& = & - \tan ^2 x \\
\end{eqnarray*}
