Question

Prove the identity

\(arcsin\frac{{x - 1}}{{x + 1}} = 2arctan\sqrt x - \frac{\pi }{2}\).

Short answer

The identity \(\arcsin \frac{{x - 1}}{{x + 1}} = 2\arctan \sqrt x - \frac{\pi }{2}\) is proved.

Step-by-step solution

 Step 1: Given data

The functions is \(\arcsin \frac{{x - 1}}{{x + 1}} = 2\arctan \sqrt x - \frac{\pi }{2}\).

Use theorem

"If\({f^\prime }(x) = 0\)for all\(x\)in an interval\((a,\;b)\), then\(f\)is constant on\((a,\;b)\)."

Find the value for \(x + 1\)

Rewrite the given equation as follows:

\({\sin ^{ - 1}}\frac{{x - 1}}{{x + 1}} = 2{\tan ^{ - 1}}\sqrt x - \frac{\pi }{2}\).

Let, \({\sin ^{ - 1}}\frac{{x - 1}}{{x + 1}} = \alpha \) and solve for \(\alpha = 2{\tan ^{ - 1}}\sqrt x - \frac{\pi }{2}\) as follows:

\(\begin{aligned}{c}{\sin ^{ - 1}}\frac{{x - 1}}{{x + 1}} &= \alpha \\\frac{{x - 1}}{{x + 1}} &= \sin \alpha \\\frac{{(x - 1) + 2 - 2}}{{x + 1}} &= \sin \alpha \\\frac{{x - 1 + 2 - 2}}{{x + 1}} &= \sin \alpha \end{aligned}\)

Simplify \(\frac{{x - 1 + 2 - 2}}{{x + 1}} = \sin \alpha \) as follows:

\(\begin{aligned}{c}\frac{{(x + 1) - 2}}{{x + 1}} &= \sin \alpha \\1 - \frac{2}{{x + 1}} &= \sin \alpha \\1 - \sin \alpha &= \frac{2}{{x + 1}}\\x + 1 &= \frac{2}{{1 - \sin \alpha }}\end{aligned}\)

Solve for \({\tan ^{ - 1}}\sqrt x \)

Simplify the above equation and obtain the value of \({\tan ^{ - 1}}\sqrt x \) as follows:

\(\begin{aligned}{c}x &= \frac{2}{{1 - \sin \alpha }} - 1x\\ &= \frac{{2 - (1 - \sin \alpha )}}{{1 - \sin \alpha }}x\\ &= \frac{{1 + \sin \alpha }}{{1 - \sin \alpha }}x\\ &= \frac{{1 - \cos \left( {\frac{\pi }{2} + \alpha } \right)}}{{1 + \cos \left( {\frac{\pi }{2} + \alpha } \right)}}\end{aligned}\)

Simplify \(x = \frac{{1 - \cos \left( {\frac{\pi }{2} + \alpha } \right)}}{{1 + \cos \left( {\frac{\pi }{2} + \alpha } \right)}}\) as follows:

Solve for \(2{\tan ^{ - 1}}\sqrt x  - \frac{\pi }{2}\)

Simplify the above equation and obtain the value of \(2{\tan ^{ - 1}}\sqrt x - \frac{\pi }{2}\) as follows:

Hence, the given identity is proved.