If x > 0, y > 0 and x > y, then $tan^{-1}\frac{x}{y}+tan^{-1}\left(\frac{x+y}{x-y}\right)$ is equal to
Answer & explanation
Correct answer: option 3
$tan^{-1}\frac{x}{y}+tan^{-1}\left(\frac{x+y}{x-y}\right)$
$=tan^{-1}\frac{x}{y}+tan^{-1}\left(\frac{1+\frac{x}{y}}{\frac{x}{y}-1}\right)$
$=tan^{-1}\frac{x}{y}-tan^{-1}\left(\frac{1+\frac{x}{y}}{1-\frac{x}{y}}\right)$
$=tan^{-1}\frac{x}{y}-\left(-\pi +tan^{-1} 1 + tan^{-1}\frac{x}{y}\right)$ $[∵\frac{x}{y} > 1 ]$
$= \pi -\frac{\pi}{4}=\frac{3\pi}{4}$