Here's the question you clicked on:
Hannah_Ahn
Prove the identity \[\left( \cot \theta \over \sin \theta - \csc \theta \right) = -\sec \theta\]
first we do the algebra \[\frac{\frac{a}{b}}{b-\frac{1}{b}}=\frac{a}{b^2-1}\]
no replace a by cosine, b by sine and get \[\frac{\cos(x)}{\sin^2(x)-1}=-\frac{\cos(x)}{1-\sin^2(x)}=-\frac{\cos(x)}{\cos^2(x)}=-\sec(x)\]
\[a \over b^2 - 1\] ?????
oooh gotcha. i thought that was the right side,
yes \[\frac{\frac{a}{b}}{b-\frac{1}{b}}=\frac{a}{b^2-1}\] multiply top and bottom by \[b\]and get it in one step