# What is Cot(arcsin (-5/13)) ?

Jul 21, 2015

$\cot \left(\arcsin \left(- \frac{5}{13}\right)\right) = - \frac{12}{5}$

#### Explanation:

Let$\text{ } \theta = \arcsin \left(- \frac{5}{13}\right)$

This means that we are now looking for color(red)cottheta!

$\implies \sin \left(\theta\right) = - \frac{5}{13}$

Use the identity,

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

NB : $\sin \theta$ is negative so $\theta$ is also negative.

We shall the importance of this info later.

$\implies \frac{{\cos}^{2} \theta + {\sin}^{2} \theta}{\sin} ^ 2 \theta = \frac{1}{\sin} ^ 2 \theta$

$\implies {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta + 1 = \frac{1}{\sin} ^ 2 \theta$

$\implies {\cot}^{2} \theta + 1 = \frac{1}{\sin} ^ 2 \theta$

$\implies {\cot}^{2} \theta = \frac{1}{\sin} ^ 2 x - 1$

$\implies \cot \theta = \pm \sqrt{\frac{1}{\sin} ^ 2 \left(\theta\right) - 1}$

$\implies \cot \theta = \pm \sqrt{\frac{1}{- \frac{5}{13}} ^ 2 - 1} = \pm \sqrt{\frac{169}{25} - 1} = \pm \sqrt{\frac{144}{25}} = \pm \frac{12}{5}$

WE saw the evidence previously that $\theta$ should be negative only.

And since $\cot \theta$ is odd $\implies \cot t \left(- A\right) = - \cot \left(A\right)$ Where $A$ is a positive angle.

So, it becomes clear that $\cot \theta = \textcolor{b l u e}{+} \frac{12}{5}$

REMEMBER what we called $\theta$ was actually $\arcsin \left(- \frac{15}{13}\right)$

$\implies \cot \left(\arcsin \left(- \frac{5}{13}\right)\right) = \textcolor{b l u e}{\frac{12}{5}}$