Question

Prove the identity
\[\left( \cot \theta \over \sin \theta - \csc \theta \right) = -\sec \theta\]

Answer 1

first we do the algebra
\[\frac{\frac{a}{b}}{b-\frac{1}{b}}=\frac{a}{b^2-1}\]

Answer 2

no replace a by cosine, b by sine and get
\[\frac{\cos(x)}{\sin^2(x)-1}=-\frac{\cos(x)}{1-\sin^2(x)}=-\frac{\cos(x)}{\cos^2(x)}=-\sec(x)\]

Answer 3

\[a \over b^2 - 1\] ?????

Answer 4

oooh gotcha. i thought that was the right side,

Answer 5

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

Answer 6

thanks! :)