Question

prove that tan^(-1) [x/y] + tan^(-1) [x-y/x+y] =pi/4​

Answer 1

(kindly note that your question has a small mistake, you should have given a minus sign instead of the plus.)

Answer:

$\tt tan^{-1} (\dfrac{x}{y} )-tan ^{-1} (\dfrac{x-y}{x+y} ) \\ \\ \\ =tan^{-1} (\dfrac{\frac{x}{y}-\frac{x-y}{x+y} }{1+(\frac{x}{y})(\frac{x-y}{x+y} ) } ) \\ \\ \\ = tan^{-1} (\dfrac{\frac{x(x+y)-y(x-y)}{y(x+y)} }{\frac{y(x+y)+x(x-y)}{y(x+y)} } ) \\ \\ \\ =tan ^{-1} (\dfrac{x^{2}+xy-xy+y^{2} }{ {xy}+ y^{2}+x^{2}-xy } ) \\ \\ \\ = tan^{-1} (\dfrac{x^{2}+y^{2} }{x^{2}+y^{2} } ) \\ \\ \\ = tan^{-1} 1 =\dfrac{\pi}{4}$

//Hence proved...

Answer 2

$\bigstar$ Question :

To prove : $\bf{tan^{-1}\left(\dfrac{x}{y}\right)+tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\dfrac{\pi }{4}}$

$\bigstar$ Answer :

$\bf{\displaystyle{{\tan}^{{-{1}}}{\left(\dfrac{x}{{y}}\right)}}-{{\tan}^{{-{1}}}{\left(\dfrac{{{x}-{y}}}{{{x}+{y}}}\right)}}}$

$\bf{\Rightarrow \displaystyle\dfrac{{{\tan}^{{-{1}}}{\left({\left(\dfrac{x}{{y}}\right)}-{\left(\dfrac{{{x}-{y}}}{{{x}+{y}}}\right)}\right)}}}{{{\left(\dfrac{x}{{y}}\right)}\cdot{\left(\dfrac{{{x}-{y}}}{{{x}+{y}}}\right)}}}}$

$\bf{\Rightarrow \displaystyle{{\tan}^{{-{1}}}{\left(\dfrac{{{x}^{2}+{x}{y}-{x}{y}+{y}^{2}}}{{{x}{y}+{y}^{2}+{x}^{2}-{x}{y}}}\right)}}}$

$\bf{\Rightarrow \displaystyle{{\tan}^{{-{1}}}{\left(\frac{{{x}^{2}+{y}^{2}}}{{{x}^{2}+{y}^{2}}}\right)}}}$

$\bf{\Rightarrow \displaystyle{{\tan}^{{-{1}}}{\left({1}\right)}}}$

$\bf{=\dfrac{\pi }{4}}$