How do you find the exact value of $tan ({sin}^{- 1} (0.1))$ $tan(sin^-1(0.1))$ ?

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How do you find the exact value of $tan ({sin}^{- 1} (0.1))$ $tan(sin^-1(0.1))$ ?

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Answer 1 · 2017-04-13T03:06:22

$\frac{1}{3 \sqrt{11}}$ $1/(3sqrt(11))$

Explanation:

Let $θ = {sin}^{- 1} (0.1) \Rightarrow sin (θ) = 0.1 = \frac{1}{10} = \frac{Opposite}{Hypotenuse}$ $theta=sin^(-1)(0.1) Rightarrow sin(theta)=0.1=1/10=("Opposite")/("Hypotenuse")$

Let $(Opposite) = 1$ $("Opposite")=1$ and $(Hypotenuse) = 10$ $("Hypotenuse")=10$ .

By Pythagorean Theorem,

$(Adjacent) = \sqrt{10^{2} - 1^{2}} = \sqrt{99} = 3 \sqrt{11}$ $("Adjacent")=sqrt(10^2-1^2)=sqrt(99)=3sqrt(11)$

Hence,

$tan ({sin}^{- 1} (0.1)) = tan θ = \frac{Opposite}{Adjacent} = \frac{1}{3 \sqrt{11}}$ $tan(sin^(-1)(0.1))=tan theta=("Opposite")/("Adjacent")=1/(3sqrt(11))$

I hope that this was clear.

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How do you find the exact value of $tan ({sin}^{- 1} (0.1))$ $tan(sin^-1(0.1))$ ?

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