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