# How do you simplify (csc² t - cot² t) / sin² t?

Mar 4, 2018

${\sec}^{2} t$

#### Explanation:

I like just sines and cosines, so let's work with those:

$\frac{{\csc}^{2} t - {\cot}^{2} t}{{\sin}^{2} \left(t\right)} = \frac{\frac{1}{{\sin}^{2} t} - \frac{{\cos}^{2} t}{{\sin}^{2} t}}{{\sin}^{2} t}$

Let's get rid of the fractions up top by multiplying top and bottom by ${\sin}^{2} t$

$= \frac{1 - {\cos}^{2} t}{{\sin}^{4} t}$

Because ${\sin}^{2} t + {\cos}^{2} t = 1$, $1 - {\cos}^{2} t = {\sin}^{2} t$.
$= \frac{{\sin}^{2} t}{{\sin}^{4} t} = \frac{1}{{\sin}^{2} t} = {\sec}^{2} t$

In hindsight, we could have also realized that there is a trig identity that
${\csc}^{2} t - {\cot}^{2} t = 1$
but that involves memorization, whereas the first thing doesn't.