Question

Solve \(tan^{-1} (\frac xy)-tan^{-1}(\frac {x-y}{x+y})\) is equal to

• A. \(\frac {\pi}{2}\)
• B. \(\frac {\pi}{3}\)
• C. \(\frac {\pi}{4}\)
• D. \(-\frac {3\pi}{2}\)

Answer 1

The Correct Option is C

Given :
\(\tan^{-1}(\frac{x}{y})-\tan^{-1}\frac{x-y}{x+y}\)
\(=\tan^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+(\frac{x}{y})(\frac{x-y}{x+y})}\right]\)

\(=\tan^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]\)
\(=\tan^{-1}\left(\frac{x^2+xy-xy+y^2}{xy+y^2+x^2-xy}\frac{}{}\right)\)
\(=\tan^{-1}\left(\frac{x^2+y^2}{x^2+y^2}\right)\)
\(=\tan^{-1}1=\frac{\pi}{4}\)
So, the correct option is (C) : \(\frac{\pi}{4}\).