# What is tan(arcsin(12/13)) ?

Jul 21, 2015

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

#### Explanation:

Let$\text{ } \theta = \arcsin \left(\frac{12}{13}\right)$

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

$\implies \sin \left(\theta\right) = \frac{12}{13}$

Use the identity,

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

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

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

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

$\implies \tan \theta = \sqrt{\frac{1}{\cos} ^ 2 \left(\theta\right) - 1}$

Recall : ${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

$\implies \tan \theta = \sqrt{\frac{1}{1 - {\sin}^{2} \theta} - 1}$

$\implies \tan \theta = \sqrt{\frac{1}{1 - {\left(\frac{12}{13}\right)}^{2}} - 1}$

=>tantheta=sqrt(169/(169-144)-1

$\implies \tan \theta = \sqrt{\frac{169}{25} - 1}$

$\implies \tan \theta = \sqrt{\frac{144}{5}} = \frac{12}{5}$

REMEMBER what we called $\theta$ was actually $\arcsin \left(\frac{12}{13}\right)$

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