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

Staff member

#### MarkFL

##### Administrator
Staff member
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:

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

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

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