Prove that:

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

Formula Used:

1) $\cot ^{-1} x=\tan ^{-1} \frac{1}{x}$

2) $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

Proof:

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

$=\tan ^{-1} x+\tan ^{-1} \frac{1}{(x+1)}$

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

$=\tan ^{-1} \frac{x(x+1)+1}{x+1-x}$

$=\tan ^{-1}\left(x^{2}+x+1\right)$

= RHS

Therefore, LHS = RHS

Hence proved.