Question
Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$

Answer

Let, $\text{x}=\text{a}\cos\theta$
Now,
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}=\tan^{-1}\sqrt{\frac{\text{a}-\text{a}\cos\theta}{\text{a}+\text{a}\cos\theta}}$
$=\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
$=\tan^{-1}\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}$
$=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)$
$=\frac{\theta}{2}$
$=\frac{1}{2}\cos^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$
$\therefore\ \tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}=\frac{\cos^{-1}\big(\frac{\text{x}}{\text{a}}\big)}{2}$

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