Find the value of \(tan^{-1}(\frac xy)-tan^{-1}(\frac{x-y}{x+y})\).

1.	Find the value of \(tan^{-1}(\frac xy)-tan^{-1}(\frac{x-y}{x+y})\).
Answer» \(tan^{-1}(\frac xy)-tan^{-1}(\frac{x-y}{x+y})\) \(=tan^{-1}\left(\cfrac{\frac xy-\frac{x-y}{x+y}}{1+\frac xy.\frac{x-y}{x+y}}\right)\) \([Here \frac xy.\frac{x-y}{x+y}>-1]\) \(=tan^{-1}(\frac{x^2+xy-xy+y^2}{y(x+y)}\times\)\(\frac{y(x+y)}{xy+y^2+x^2-xy})\) \(=tan^{-1}(\frac{x^2+y^2}{x^2+y^2})\) \(=tan^{-1}(1)=\frac{\pi}4\)

Answer»

\(tan^{-1}(\frac xy)-tan^{-1}(\frac{x-y}{x+y})\)

\(=tan^{-1}\left(\cfrac{\frac xy-\frac{x-y}{x+y}}{1+\frac xy.\frac{x-y}{x+y}}\right)\) \([Here \frac xy.\frac{x-y}{x+y}>-1]\)

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

\(=tan^{-1}(\frac{x^2+y^2}{x^2+y^2})\)

\(=tan^{-1}(1)=\frac{\pi}4\)

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