How do you simplify $tan ({sin}^{- 1} (x))$ $tan(sin^-1(x))$ ?

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How do you simplify $tan ({sin}^{- 1} (x))$ $tan(sin^-1(x))$ ?

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Answer 1 · 2016-05-27T16:57:04

Let ${sin}^{- 1} x = θ$ $sin^-1x=theta$ hence $x = sin θ$ $x=sintheta$

For $0 < x < 1$ $0<x<1$ we draw a right triangle with hypotenuse equal to 1 and the other side equals to $x$ $x$ like the one in the Figure below.
From pythagorean theorem the other side is $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$

Now we know that

$tan θ = \frac{sin θ}{cos θ} = \frac{sin θ}{\sqrt{1 - {sin}^{2} θ}}$ $tantheta=(sintheta)/(costheta)=(sintheta)/(sqrt(1-sin^2theta))$

Because $x = sin θ$ $x=sintheta$

We have that

$tan θ = \frac{x}{\sqrt{1 - x^{2}}}$ $tantheta=x/(sqrt(1-x^2))$

But from ${sin}^{- 1} x = θ$ $sin^-1x=theta$ we get

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

Figure

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How do you simplify $tan ({sin}^{- 1} (x))$ $tan(sin^-1(x))$ ?

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