# How do you solve cot [Arcsin (-12/13)]?

Jun 28, 2016

$- \frac{5}{12}$ against the principal value of $a r c \sin \left(- \frac{123}{13}\right)$ and, for the general value, the answer is $\pm \frac{5}{12}$.

#### Explanation:

Let $a = a r c \sin \left(- \frac{12}{13}\right)$. Then, $\sin a = - \frac{12}{13} < 0$. The principal

value of a is in the 4th quadrant. The general value is either in the

4th or in the 3rd.. So. cos a is $\sqrt{1 - {12}^{2} / {13}^{2}} = \frac{5}{13}$ or $\pm \frac{5}{13}$.

The given expression is cot a = cos a/sin a and, accordingly, the