Question

Find the value of ${{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)$

Answer 1

Type of question is based on the trigonometric identities. As it looks somewhat complicated but if we know the identities and their use very well then these types of questions become easy for us. We should have an aim of making it as simple as possible to these types of questions through the identities we know.

Complete step-by-step solution:
As this question looks like ${{\tan }^{-1}}a-{{\tan }^{-1}}b$ of which we know the identity i.e. ${{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$i.e. ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$; So moving ahead with the question we will apply the ${{\tan }^{-1}}a-{{\tan }^{-1}}b$ identity to simplify the question through which we can easily get the result.
By comparing the value of ${{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)$ with ${{\tan }^{-1}}a-{{\tan }^{-1}}b$ we will get $a=\dfrac{x}{y}$and $b=\dfrac{x-y}{x+y}$. So we will get;

& ={{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right) \\\ & As, \\\ & {{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right) \\\ \end{aligned}$$ By comparing both equation, we will get; $$\begin{aligned} & {{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{x}{y}-\dfrac{x-y}{x+y}}{1+\left( \dfrac{x}{y} \right)\left( \dfrac{x-y}{x+y} \right)} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{\dfrac{x\left( x+y \right)-\left( x-y \right)y}{y\left( x+y \right)}}{\dfrac{y(x+y)+x(x-y)}{y\left( x+y \right)}} \right) \\\ & ={{\tan }^{-1}}\left( \dfrac{{{x}^{2}}+xy-xy+{{y}^{2}}}{{{x}^{2}}+xy-xy+{{y}^{2}}} \right) \\\ \end{aligned}$$ As both numerator and denominator have same expressions so they will get cancelled and we will get 1, i.e. $$\begin{aligned} & ={{\tan }^{-1}}1 \\\ & =\dfrac{\pi }{4} \\\ \end{aligned}$$ And as we know that $${{\tan }^{-1}}1$$ is equal to $$\dfrac{\pi }{4}$$. **Hence the answer is $$\dfrac{\pi }{4}$$, so the value of ${{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)$ is equal to $$\dfrac{\pi }{4}$$** **Note:** In order to solve such questions we are required to have a practice on recognising which identity to be used and how to use them. Moreover, being patient while using identity as use of sign is a general mistake we make while solving trigonometry. Here we use the identity of the inverse trigonometric function.