Question

Evaluate ${\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(\frac{x-y}{x+y}\right)$ is

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{4}\text{or -}\frac{3\mathrm{\pi }}{4}$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Find the value of the given Equation.

Consider the given Equation as,

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

Rewrite the above Equation by dividing the second term of the Equation by $y$ as,

$\Rightarrow {\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(\frac{{\displaystyle \frac{x-y}{y}}}{{\displaystyle \frac{x+y}{y}}}\right)={\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(\frac{{\displaystyle \frac{x}{y}-1}}{1+{\displaystyle \frac{x}{y}}}\right)$$$\begin{gathered} ={\tan }^{-1}\left(\frac{x}{y}\right)-\left[{\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(1\right)\right]\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{A}\mathbf{-}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{B}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{{\displaystyle A-B}}{1+A.B}\right)\mathbf{]} \\ ={\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(1\right) \\ ={\tan }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{4}\mathbf{}\left[\because {\tan }^{-1}\left(1\right)=\frac{\pi }{4}\right] \end{gathered}$$

Therefore,

${\tan }^{-1}\left(\frac{x}{y}\right)-{\tan }^{-1}\left(\frac{x-y}{x+y}\right)=\frac{\mathrm{\pi }}{4}$

Hence, option (C) is the correct answer.