How do you solve and find the value of $tan ({sin}^{- 1} (\frac{1}{2}))$ $tan(sin^-1(1/2))$ ?

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How do you solve and find the value of $tan ({sin}^{- 1} (\frac{1}{2}))$ $tan(sin^-1(1/2))$ ?

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Answer 1 · 2016-11-26T06:09:02

$tan ({sin}^{- 1} (\frac{1}{2})) = \frac{1}{\sqrt{3}}$ $tan(sin^(-1)(1/2))=1/sqrt3$

Explanation:

Inverse trigonometric ratios are the inverse functions of the trigonometric functions. As an example, $y = sin x$ $y=sinx$ means sine of $∠ x$ $/_x$ is $y$ $y$ . As an inverse trigonometric ration this means that $arcsin y = x$ $arcsiny=x$ or ${sin}^{- 1} y = x$ $sin^(-1)y=x$ .

As $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$ , we have ${sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$ $sin^(-1)(1/2)=pi/6$

and hence

$tan ({sin}^{- 1} (\frac{1}{2})) = tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}$ $tan(sin^(-1)(1/2))=tan(pi/6)=1/sqrt3$

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How do you solve and find the value of $tan ({sin}^{- 1} (\frac{1}{2}))$ $tan(sin^-1(1/2))$ ?

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