# How do you simplify csc θsec θ - cot θ?

Nov 12, 2015

$\tan \theta$

#### Explanation:

Lets start by using some trig identities to put the original statement into terms of $\sin$ and $\cos$. Substituting the trig identities for $\sec$, $\csc$, and $\cot$ the expression becomes;

$\frac{1}{\sin} \theta \frac{1}{\cos} \theta - \cos \frac{\theta}{\sin} \theta$

At this point it would be helpful to have a common denominator. We can do that by multiplying the top and bottom of the second term by $\cos \theta$.

$\frac{1}{\sin} \theta \frac{1}{\cos} \theta - \cos \frac{\theta}{\sin} \theta \cos \frac{\theta}{\cos} \theta$

Now we can combine the numerators over $\sin \theta \cos \theta$.

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

We can use the Pythagorean theorem to simplify the numerator. The Pythagorean theorem states that;

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

Rearranging the terms we get;

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

Now we can make the substitution for the numerator.

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

The $\sin \theta$ terms cancel, leaving;

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

Which is the definition of $\tan$.

$\tan \theta$