Question

Last Question!

sin^2x -1/ cos(-x)

Answer 1

\[\frac{\sin^2x-1}{\cos(-x)}= \frac{-(1-\sin^2x)}{cosx}=\frac{-\cos^2x}{cosx} = -cosx\]

Answer 2

Wow that's really good. Thank you!

Answer 3

since \[\cos(-\theta) = \cos \theta \rightarrow \cos(-x) = cosx\]