Question

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

Answer 1

\[\large\rm \frac{\sin^2x-1}{\cos(-x)}\]Cosine is an `even function`, meaning \(\large\rm \cos(-x)=\cos x\)

Answer 2

After making that change,
apply your Pythagorean Identity in the numerator,
\[\large\rm 1-\sin^2x=\cos^2x\]Oh I guess we need to reverse the terms in the numerator to be able to apply this identity,\[\large\rm \frac{\sin^2x-1}{\cos(-x)}\quad=\quad \frac{-(1-\sin^2x)}{\cos(-x)}\]Factoring a -1 out will do it for us.

Answer 3

-cosx/cosx then?

Answer 4

-(cosx^2)/cosx so -cosx(cosx)/cosx=-cosx?

Answer 5

yayyy good job!