# How do you calculate Tan (sin^-1 (2/3)) ?

Aug 29, 2015

$\frac{2}{\sqrt{5}}$

#### Explanation:

Drawing the right angled triangle, you realise that length of opposite side $= 2$ and length of hypotenuse $= 3 \setminus R i g h t a r r o w$ length of adjacent side $= \sqrt{{3}^{2} - {2}^{2}} = \sqrt{5}$

Thus $\tan \left({\sin}^{-} 1 \left(\frac{2}{3}\right)\right) = \frac{o p p o s i t e}{a \mathrm{dj} a c e n t} = \frac{2}{\sqrt{5}}$

Aug 29, 2015

$\tan \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right) = \pm \frac{2}{5} \sqrt{5}$

#### Explanation:

If $x = {\sin}^{- 1} \left(\frac{2}{3}\right)$
then $\sin \left(x\right) = \frac{2}{3}$

And we have one of the defining triangles below:
from which it follows:
$\tan \left({\sin}^{- 1} \left(\frac{2}{3}\right)\right) = \pm \frac{2}{\sqrt{5}} = \pm \frac{2}{5} \sqrt{5}$