First slide

Properties of ITF

Question

If $x > 0, y > 0 and x > y$ then $\tan^{- 1} \frac{x}{y} + \tan^{- 1} (\frac{x + y}{x - y})$ is equal to

Moderate

A

$- \frac{π}{4}$
B

$\frac{π}{4}$
C

$\frac{3 π}{4}$
D

none of these

Solution

$\tan^{- 1} \frac{x}{y} + \tan^{- 1} (\frac{x + y}{x - y})$
$\begin{array}{l} = \tan^{- 1} \frac{x}{y} + \tan^{- 1} (\frac{1 + \frac{x}{y}}{\frac{x}{y} - 1}) \\ = \tan^{- 1} \frac{x}{y} - \tan^{- 1} (\frac{1 + \frac{x}{y}}{1 - \frac{x}{y}}) \\ = \tan^{- 1} \frac{x}{y} - (- π + \tan^{- 1} 1 + \tan^{- 1} \frac{x}{y}) [∵ \frac{x}{y} > 1] \\ = π - \frac{π}{4} = \frac{3 π}{4} \end{array}$

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