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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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