Question

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

Answer 1

$L.H.S={\tan }^{-1}x+{\cot }^{-1}(x+1){\cot }^{-1}(x+1)={\tan }^{-1}\frac{1}{(x+1)}\therefore {\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{x+1}\right)={\tan }^{-1}\left[\frac{x+\frac{1}{x+1}}{1-\frac{x}{x+1}}\right]{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left[\frac{A+B}{1-AB}\right]={\tan }^{-1}[\frac{\left[x\right(x+1)+1]}{(x+1)}\times \frac{(x+1)}{\left[\right(x+1)-x]}]={\tan }^{-1}\left[\frac{x^2+x+1}{1}\right]=R.H.S\therefore L.H.S=R.H.S$