Question

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

Answer 1

Hint: Use the formula for difference of two tan inverse functions, given by, ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$. Simplify the numerator and denominator and cancel the common terms to get a numerical value. Now, find the angle whose, when tangent is taken, equals to the numerical value obtained by the simplification in the first step. Use the formula: ${{\tan }^{-1}}\left( \tan x \right)=x$ where $x \in ({\dfrac{-\pi}{2},\dfrac{\pi}{2}})$.

Complete step-by-step solution -
We have been given: ${{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)$.
Now, applying the formula for difference of two tan inverse functions, given by, ${{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a-b}{1+ab} \right)$, we get,
$\begin{align}
  & {{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{x}{y}-\dfrac{x-y}{x+y}}{1+\dfrac{x}{y}\times \left( \dfrac{x-y}{x+y} \right)} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{x}{y}-\dfrac{x-y}{x+y}}{1+\left( \dfrac{{{x}^{2}}-xy}{xy+{{y}^{2}}} \right)} \right) \\
\end{align}$
Now, taking L.C.M separately in the numerator and denominator, we get the expression
\[\begin{align}
  & ={{\tan }^{-1}}\left( \dfrac{\dfrac{x\left( x+y \right)-y\left( x-y \right)}{xy+{{y}^{2}}}}{\dfrac{xy+{{y}^{2}}+x-x{{y}^{2}}}{xy+{{y}^{2}}}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{{{x}^{2}}+xy-xy+{{y}^{2}}}{xy+{{y}^{2}}}}{\dfrac{xy+{{y}^{2}}+{{x}^{2}}-xy}{xy+{{y}^{2}}}} \right) \\
 & ={{\tan }^{-1}}\left( \dfrac{\dfrac{{{x}^{2}}+{{y}^{2}}}{xy+{{y}^{2}}}}{\dfrac{{{y}^{2}}+{{x}^{2}}}{xy+{{y}^{2}}}} \right) \\
\end{align}\]
Cancelling the common terms, we get the expression
$={{\tan }^{-1}}\left( 1 \right)$
Now, we have to find such an angle whose tangent is one and the angle must lie in the range $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.
We know that, $\tan \dfrac{\pi }{4}=1$. So, taking inverse function both sides, we get,
${{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)={{\tan}^{-1}} {\left( 1 \right)}$
Now, using the identity, ${{\tan }^{-1}}\left( \tan x \right)=x$ where $x \in ({\dfrac{-\pi}{2},\dfrac{\pi}{2}})$, we get
$\begin{align}
  & \dfrac{\pi }{4}={{\tan }^{-1}}\left( 1 \right) \\
 & \Rightarrow {{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4} \\
\end{align}$
Therefore, the value of the expression: ${{\tan }^{-1}}\left( \dfrac{x}{y} \right)-{{\tan }^{-1}}\left( \dfrac{x-y}{x+y} \right)$ is 1.

Note: One may note that there is only one principal value of an inverse trigonometric function. We know that, at many angles the value of tan is 1 but we have to remember the range in which tan inverse function is defined. We have to choose such an angle which lies in the range and satisfies the function. So, there can be only one answer.