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

Solution

$\tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x - y}{x + y}) = \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{\frac{x}{y} - 1}{\frac{x}{y} + 1}) = \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{\frac{x}{y} - 1}{1 + \frac{x}{y}}) = \tan^{- 1} (\frac{x}{y}) - [\tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (1)] [∵ \tan^{- 1} a - \tan^{- 1} b = \tan^{- 1} (\frac{a - b}{1 + ab})]$

= $= \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x}{y}) + \tan^{- 1} (1) = \tan^{- 1} (1) = \frac{π}{4} Thus, \tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x - y}{x + y}) = \frac{π}{4}$

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

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