# How do you find the exact value of tan (sin^-1 (2/3)) ?

Since $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$, we must find $\sin \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right)$ and $\cos \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right)$. Clearly, $\sin \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right) = \frac{2}{3}$ by the definition of an inverse function (in this case, to be more precise, $\sin \left({\sin}^{- 1} \left(x\right)\right) = x$ for all $- 1 \setminus \le q x \setminus \le q 1$). Also, the cosine of the "angle" ${\sin}^{- 1} \left(\frac{2}{3}\right)$ is positive, therefore by the Pythagorean identity, $\cos \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right) = \sqrt{1 - {\sin}^{2} \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right)}$
$= \sqrt{1 - {\left(\frac{2}{3}\right)}^{2}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3.}$
(You could also draw a right triangle, label one of the angles ${\sin}^{- 1} \left(\frac{2}{3}\right)$, label the side lengths appropriately, use the Pythagorean theorem and SOH, CAH, TOA appropriately to do this last calculation.)
Therefore, $\tan \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right) = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$