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Chapter 2: Problem 151

If \(\alpha=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right)\) and \(\beta=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) for \(0

Short Answer

Expert verified
\(\alpha+\beta=\pi\)

Step by step solution

01

Simplify the Equation for \(\alpha\)

Utilize the identity \(2 \tan ^{-1}(t)=\pi - 2 \tan ^{-1}(\frac{1}{t})\). Plugging \(t= \frac{1 + x}{1 - x}\) into the identity, the equation for \(\alpha\) simplifies to \(\alpha = \pi - 2 \tan ^{-1}(1-x)\).
02

Simplify the Equation for \(\beta\)

Utilize the identity \(\sin ^{-1}(t)= \pi/2 - \cos ^{-1}(t)\). Plugging \(t= \frac{1 - x^{2}}{1 + x^{2}}\) into the identity, the equation for \(\beta\) simplifies to \(\beta = \pi/2 - \cos^{-1}\left(\frac{2x}{1+x^2}\right)\). As per the well-known identity, we have \(\cos ^{-1}(t)=2 \tan ^{-1}\left( \sqrt{1-t^2}/(1+t)\right)\). Substituting \(t=2x/(1+x^2)\) into this identity we have \(\cos ^{-1}\left(\frac{2x}{1 + x^{2}}\right) = 2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)\). Therefore, \(\beta = \pi/2 - 2 \tan ^{-1}(\frac{1-x}{1+x})\).
03

Add \(\alpha\) and \(\beta\)

The expression for \(\alpha + \beta\) becomes: \(\alpha + \beta = \pi - 2 \tan ^{-1}(1-x) + \pi/2 - 2 \tan ^{-1}(\frac{1-x}{1+x})\). Notice how \(2 \tan ^{-1}(1-x) + 2 \tan ^{-1}\left(\frac{1-x}{1+x}\right) = \pi/2\). Therefore, \(\alpha + \beta = \pi + \pi/2 - \pi/2 = \pi\).

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