Question

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

• A. ${\tan }^{-1}x$
• B. ${\cot }^{-1}x$
• C. ${\tan }^{-1}y$
• D. ${\cot }^{-1}y$

Answer 1

Answer: (B)

${\tan }^{-1}\left(\frac{1}{x+y}\right)+{\tan }^{-1}\left(\frac{y}{x^2+xy+1}\right)$
$={\tan }^{-1}\left[\frac{\frac{1}{x+y}+\frac{y}{x^2+xy+1}}{1-\frac{1}{\left(x+y\right)}\frac{y}{\left(x^2+xy+1\right)}}\right]$
$={\tan }^{-1}\left[\frac{x^2+xy+1+xy+y^2}{x^3+x^{2}y+x+y{x}^2+x{y}^2+y-y}\right]$
$={\tan }^{-1}\left[\frac{x^2+2xy+y^2+1}{x\left(x^2+2xy+y^2+1\right)}\right]$
$={\tan }^{-1}\frac{1}{x}={\cot }^{-1}x$