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.

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