Prove that:

Solution:

To Prove: $\cot ^{-1}\left(\sqrt{1+x^{2}}-x\right)=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$

Formula Used:

1) $\tan \left(\frac{\pi}{4}+\mathrm{A}\right)=\frac{1+\tan \mathrm{A}}{1-\tan \mathrm{A}}$

2) $\operatorname{cosec}^{2} A=1+\cot ^{2} A$

3) $1-\cos A=2 \sin ^{2}\left(\frac{A}{2}\right)$

4) $\sin A=2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)$

Proof:

$\mathrm{LHS}=\cot ^{-1}\left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)$

Let $x=\cot A$

$\mathrm{LHS}=\cot ^{-1}\left(\sqrt{1+\cot ^{2} \mathrm{~A}}-\cot \mathrm{A}\right)$

$=\cot ^{-1}(\operatorname{cosec} A-\cot A)$

$=\cot ^{-1}\left(\frac{1-\cos \mathrm{A}}{\sin \mathrm{A}}\right)$

$=\cot ^{-1}\left(\frac{2 \sin ^{2}\left(\frac{\mathrm{A}}{2}\right)}{2 \sin \left(\frac{\mathrm{A}}{2}\right) \cos \left(\frac{\mathrm{A}}{2}\right)}\right)$

$=\cot ^{-1}\left(\tan \left(\frac{\mathrm{A}}{2}\right)\right)$

$=\frac{\pi}{2}-\tan ^{-1}\left(\tan \left(\frac{\mathrm{A}}{2}\right)\right)$

$=\frac{\pi}{2}-\frac{\mathrm{A}}{2}$

From $(2), A=\cot ^{-1} x$

$\frac{\pi}{2}-\frac{A}{2}=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$

$=\mathrm{RHS}$

Therefore, LHS = RHS

Hence proved.