Question

sin^2x-cos^2x/1-cot^2x

Answer 1

Okay, I'm guessing it's actually (sin^2x-cos^2x)/(1-cot^2x)...

First of all, let's write \[1-\cot^2x\] in terms of sin and cos:

\[1 - \cot^2x =\]
\[1 - (\cos^2x/\sin^2x) =\]
\[(\sin^2x / \sin^2x) - (\cos^2x) / (\sin^2x) = \]
\[(\sin^2x - \cos^2x) / \sin^2x\]

Well, that's nifty, because now I've got something similar on the top and the bottom.
\[(\sin^2x - \cos^2x) / ((\sin^2x - \cos^2x) / \sin^2x) = \]
\[((\sin^2x - \cos^2x) / 1) * (\sin^2x / (\sin^2x - \cos^2x)) = \]
\[1 / \sin^2x =\]
\[\csc^2x\]