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Rebekah's question at Yahoo! Answers involving inverse trigonometric functions

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Rebekah,

We are given to evaulate:

$\displaystyle \sin^{-1}(x)+\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x} \right)$

where $0<x\le1$

Let's draw a diagram of a right triangle where:

$\displaystyle \alpha=\sin^{-1}(x)\text{ and }\beta=\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x} \right)$:

rebekah.jpg

Now, it is easy to see that $\alpha$ and $\beta$ are complementary, hence:

$\displaystyle \alpha+\beta=\sin^{-1}(x)+\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x} \right)=\frac{\pi}{2}$