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simplify the expression sin^2 x-1/ cos(-x)
the answer choices are 
-sin x 
cos x
sin x
-cos x

Sagot :

Naǫidn
[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]
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