Answer :
[tex]\hbox{using trigonometric identities:} \\ \sin^2x +\cos^2x=1 \\ \sin^2x=1-\cos^2x \\ and \\ \cos (-x)=\cos x \\ \\ \frac{\sin^2x-1}{\cos (-x)}=\frac{1-\cos^2 x-1}{\cos x}=\frac{-\cos^2x}{\cos x}=\boxed{- \cos x} \\
(x \not= \frac{\pi}{2}+n\pi, n \in \mathbb{Z})[/tex]
