Question:
Prove that:
$\tan ^{-1} x+\cot ^{-1}(x+1)=\tan ^{-1}\left(x^{2}+x+1\right)$
Solution:
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.
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