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### Answers (1)

Answer:
$\frac{\pi }{4}$
Hint:
For solving this, we can use the formula of union trigonometric function,
$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left ( \frac{a-b}{1+ab} \right )$
Given:
$\tan^{-1}\left ( \frac{x}{y} \right )-\tan^{-1}\left ( \frac{x-y}{x+y} \right )$
Explanation:
Let’s use the formula
Where$a=\frac{x}{y}\: and \: b=\frac{x-y}{x+y}$
$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)=\tan ^{-1}\left(\frac{\frac{x}{y}-\left(\frac{x-y}{x+y}\right)}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right)$
\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{x^{2}+x y-y(x-y)}{y(x+y)}}{\frac{x y+y^{2}+x^{2}-x y}{y(x+y)}}\right) \\ &=\tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}
Note: We must know the formula of union trigonometric function.

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