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What is tan $\left(\cos ^{-1} x\right)$ equal to?
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Verified Answer
The correct answer is:
$\frac{\sqrt{1-x^{2}}}{x}$
Let $\cos ^{-1} \mathrm{x}=\theta$
$\Rightarrow \cos \theta=\mathrm{x} \Rightarrow \sin \theta=\sqrt{1-\mathrm{x}^{2}}$
$\Rightarrow \tan \theta=\frac{\sqrt{1-x^{2}}}{x}$ and $\theta=\cos ^{-1} x$
This can be represented by a triangle with hypotenuous $=1$ and sides $\mathrm{x}$ and $\sqrt{1-\mathrm{x}^{2}}$.
$\Rightarrow \tan \left(\cos ^{-1} x\right)=\frac{\sqrt{1-x^{2}}}{x}$
$\Rightarrow \cos \theta=\mathrm{x} \Rightarrow \sin \theta=\sqrt{1-\mathrm{x}^{2}}$
$\Rightarrow \tan \theta=\frac{\sqrt{1-x^{2}}}{x}$ and $\theta=\cos ^{-1} x$
This can be represented by a triangle with hypotenuous $=1$ and sides $\mathrm{x}$ and $\sqrt{1-\mathrm{x}^{2}}$.
$\Rightarrow \tan \left(\cos ^{-1} x\right)=\frac{\sqrt{1-x^{2}}}{x}$
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