How do you calculate $tan ({sin}^{- 1} (\frac{2}{3}))$ $Tan (sin^-1 (2/3))$ ?

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How do you calculate $tan ({sin}^{- 1} (\frac{2}{3}))$ $Tan (sin^-1 (2/3))$ ?

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Answer 1 · 2015-08-29T17:09:19

$\frac{2}{\sqrt{5}}$ $2/sqrt(5)$

Explanation:

Drawing the right angled triangle, you realise that length of opposite side $= 2$ $=2$ and length of hypotenuse $= 3 \Rightarrow$ $=3 \Rightarrow$ length of adjacent side $= \sqrt{3^{2} - 2^{2}} = \sqrt{5}$ $= sqrt(3^2-2^2)=sqrt(5)$

Thus $tan ({sin}^{- 1} (\frac{2}{3})) = \frac{o p p o s i t e}{a d j a c e n t} = \frac{2}{\sqrt{5}}$ $tan (sin^-1 (2/3))=(opposite)/(adjacent)=2/sqrt(5)$

Answer 2 · 2015-08-29T17:21:27

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

Explanation:

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

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

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How do you calculate $tan ({sin}^{- 1} (\frac{2}{3}))$ $Tan (sin^-1 (2/3))$ ?

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Explanation:

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