# How do you solve and find the value of cot(sin^-1(5/6))?

Jan 19, 2017

$\cot \theta = + \frac{\sqrt{11}}{5}$.

#### Explanation:

Let ${\sin}^{-} 1 \left(\frac{5}{6}\right) = \theta \Rightarrow \sin \theta = \frac{5}{6} , \theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

But as $\sin \theta > 0 , \theta \in \left(0 , \frac{\pi}{2}\right)$

We require the value=$\cot \theta$

Knowing that #csc^2theta=1+cot^2theta, we have,

${\cot}^{2} \theta = {\csc}^{2} \theta - 1 = \frac{1}{\sin} ^ 2 \theta - 1 = \frac{36}{25} - 1 = \frac{11}{25}$

$\Rightarrow \cot \theta = \pm \frac{\sqrt{11}}{5}$

As, $\theta \in \left(0 , \frac{\pi}{2}\right) , \cot \theta = + \frac{\sqrt{11}}{5}$.