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

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The problem involves evaluating the expression sin^(-1)(x) + tan^(-1)(√(1-x^2)/x) for 0 < x ≤ 1. A right triangle is used to illustrate the relationship between the angles, where α = sin^(-1)(x) and β = tan^(-1)(√(1-x^2)/x). It is established that these angles are complementary, leading to the conclusion that α + β equals π/2. Thus, the final result of the expression is π/2. This solution effectively demonstrates the connection between inverse trigonometric functions and their geometric interpretations.
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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:

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

View attachment 616

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

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

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