## anonymous 4 years ago Prove the identity $\left( \cot \theta \over \sin \theta - \csc \theta \right) = -\sec \theta$

1. anonymous

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

2. anonymous

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

3. anonymous

$a \over b^2 - 1$ ?????

4. anonymous

oooh gotcha. i thought that was the right side,

5. anonymous

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

6. anonymous

thanks! :)