# What is the exact value of cot(-90) and csc690, using unit circle?

##### 1 Answer
Oct 23, 2017

It depends on what you do want to use.

#### Explanation:

I assume that you actually mean to ask about $\cot \left(- {90}^{\circ}\right)$

(There is no nice way to express the exact value of $\cot \left(- 90\right)$).

If you know the following, then there is no need to use the unit circle:

$\cot A = \cos \frac{A}{\sin} A$

$\sin \left(- A\right) = - \sin \left(A\right)$ and $\cos \left(- A\right) = \cos \left(A\right)$

$\sin \left({90}^{\circ}\right) = 1$ and $\cos \left({90}^{\circ}\right) = 0$

We get
$\cot \left({90}^{\circ}\right) = \cos \frac{- {90}^{\circ}}{\sin} \left(- {90}^{\circ}\right) = \cos \frac{{90}^{\circ}}{- \sin \left({90}^{\circ}\right)} = \frac{0}{- \left(1\right)} = 0$

For the other we can use

$\csc \left({690}^{\circ}\right) = \csc \left({690}^{\circ} - {360}^{\circ}\right) = \csc \left({330}^{\circ}\right)$

$= \csc \left({330}^{\circ} - {360}^{\circ}\right) = \csc \left(- {30}^{\circ}\right)$

$= \frac{1}{\sin} \left(- {30}^{\circ}\right) = \frac{1}{- \sin \left({30}^{\circ}\right)} = \frac{1}{- \left(\frac{1}{2}\right)} = - 2$