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Found in: Page 216

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Prove the identity$$arcsin\frac{{x - 1}}{{x + 1}} = 2arctan\sqrt x - \frac{\pi }{2}$$.

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

See the step by step solution

## Step 1: Given data

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

## Step 2: Use theorem

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

## Step 3: 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}

## Step 4: 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:

## Step 5: 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.