# How do you express tan theta - cot theta  in terms of cos theta ?

Mar 25, 2016

We must use the quotient identities, $\tan \theta = \sin \frac{\theta}{\cos} \theta$ and $\cot \theta = \cos \frac{\theta}{\sin} \theta$

#### Explanation:

= $\sin \frac{\theta}{\cos} \theta - \cos \frac{\theta}{\sin} \theta$

=${\sin}^{2} \frac{\theta}{\cos \theta \sin \theta} - {\cos}^{2} \frac{\theta}{\cos \theta \sin \theta}$

Use the pythagorean identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ to simplify further.

$= \frac{1 - {\cos}^{2} \theta - {\cos}^{2} \theta}{\cos \theta \sin \theta}$

$= \frac{1 - 2 {\cos}^{2} \theta}{\cos \theta \sin \theta}$

This is simplest form; we can't get rid of the sin (pardon the unintended pun!).

Hopefully this helps!

Mar 28, 2016

$\frac{\sqrt{1 - {\cos}^{2} \theta}}{\cos} \theta - \cos \frac{\theta}{\sqrt{1 - {\cos}^{2} \theta}}$

#### Explanation:

Express first in terms of $\sin \theta$ and $\cos \theta$.

$= \sin \frac{\theta}{\cos} \theta - \cos \frac{\theta}{\sin} \theta$

Now, to turn the $\sin \theta$ terms into $\cos \theta$ terms, use the Pythagorean identity:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

${\sin}^{2} \theta = 1 - {\cos}^{2} \theta$

$\sin \theta = \sqrt{1 - {\cos}^{2} \theta}$

Substitute this into the expression we originally made:

$= \frac{\sqrt{1 - {\cos}^{2} \theta}}{\cos} \theta - \cos \frac{\theta}{\sqrt{1 - {\cos}^{2} \theta}}$