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

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−π4

π4

3π4

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detailed solution

Correct option is C

tan−1⁡xy+tan−1⁡x+yx−y=tan−1⁡xy+tan−1⁡1+xyxy−1=tan−1⁡xy−tan−1⁡1+xy1−xy=tan−1⁡xy−−π+tan−1⁡1+tan−1⁡xy ∵xy>1=π−π4=3π4

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If x>0,y>0 and x>y then tan−1⁡xy+tan−1⁡x+yx−y is equal to

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