Question
Write the following in the simplest form:
$\tan^{-1}\Big\{\sqrt{1+\text{x}^2}-\text{x}\Big\},\text{x}\in\text{R}$

Answer

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

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