# How do you simplify the expression cot(arcsin (-7/13))?

Jun 24, 2016

$- \frac{\sqrt{120}}{7}$ against principal value of arc sin (-7/13), in the 4th quadrant. Of course, the general solution is $\pm \frac{\sqrt{120}}{7}$.

#### Explanation:

Let $a = a r c \sin \left(- \frac{7}{13}\right)$. Then, $\sin a = - \frac{7}{13} < 0$.

a is in either 3rd quadrant or in the fourth. The principal value is in

4th. Accordingly,

$\cos a = \sqrt{1 - {7}^{2} / {13}^{2}} = \frac{\sqrt{120}}{13}$, against principal a.

Also, for the 3rd quadrant a, $\cos a = - \frac{\sqrt{120}}{13}$.

The given expression is

$\cot a = \cos \frac{a}{\sin} a = - \frac{\sqrt{120}}{7}$, against principal a and

$\frac{\sqrt{120}}{7}$, for the other a.