Prove that:

Solution:

To Prove: $\tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi}{4}+\tan ^{-1} x$

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

Proof:

$\mathrm{LHS}=\tan ^{-1}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right) \ldots$ (1)

Let $x=\tan A \ldots$ (2)

Substituting (2) in (1),

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

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

$=\frac{\pi}{4}+\mathrm{A}$

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

$\frac{\pi}{4}+\mathrm{A}=\frac{\pi}{4}+\tan ^{-1} \mathrm{x}$

= RHS

Therefore, LHS = RHS

Hence proved.