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)

$\cot^{-1} x$

Answer 2

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