Browse Questions

# Prove the following : $tan^{-1} \bigg( \frac{x}{y} \bigg)- tan^{-1} \bigg( \frac{x-y}{x+y} \bigg)= \frac{\pi}{4},y \neq 0.$

This is Q.No. 17 of Misc. chapter 2

Toolbox:
• $tan^{-1}\alpha-tan^{-1}\beta=tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta}\:\:\alpha\beta<1$
Ans: (C) $\frac{\pi}{4}$
Given $tan^{-1}\large \frac {x}{y} -tan^{-1}\large \frac {x-y}{x+y}$
We know that $tan^{-1}\alpha-tan^{-1}\beta=tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta}\:\:\alpha\beta<1$
Let's take $\alpha=\large\frac{x}{y}\:and\:\beta=\large\frac{x-y}{x+y}$
$1+ \alpha \beta=1+ \large \frac{x}{y} \frac{x-y}{x+y}= \large\frac{y(x+y)+x(x-y)}{y(x+y)} = \large\frac{xy+y^2+x^2-xy}{y(x+y)} = \large\frac{x^2+y^2}{y(x+y)}$
$\alpha -\beta=\large \frac{x}{y}-\large \frac{x-y}{x+y}=\large\frac{x(x+y)+y(x-y)}{y(x+y)} =\large \frac{x^2+xy-xy+y^2}{y(x+y)} = \large\frac{x^2+y^2}{y(x+y)}$
Since $\alpha - \beta = 1+\alpha\;\beta \rightarrow tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta} = \tan^{-1}1 = \large\frac{\pi}{4}$