# How do you solve and find the value of cot(sin^-1(7/9))?

Mar 10, 2017

$\pm \frac{4 \sqrt{2}}{7}$

#### Explanation:

$\sin x = \frac{7}{9}$
cot (arcsin x) = cot x
Use trig identity:
$1 + {\cot}^{2} x = \frac{1}{{\sin}^{2} x} = \frac{81}{49}$
${\cot}^{2} x = \frac{81}{49} - 1 = \frac{32}{49}$
$\cot x = \pm \frac{4 \sqrt{2}}{7}$.
$\sin x = \frac{7}{9}$ --> x could be in Quadrant 1 or Quadrant 2. There fore,
cot x could be either negative or positive.