Let $x, y, z$ and $r$ be positive real numbers such that $x^{2} + y^{2} + z^{2} = r^{2} .$ Then the value of ${tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + tan (\frac{z x}{y r})$ is

π

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\frac{π}{2}

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0

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\frac{3 π}{2}

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Solution

The correct option is D $\frac{3 π}{2}$
We observe $\frac{x y}{z r} \frac{y z}{x r} = \frac{y^{2}}{r^{2}} = \frac{y^{2}}{x^{2} + y^{2} + z^{2}} < 1$
So, ${tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + {tan}^{- 1} (\frac{x z}{y r})$
$= {tan}^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{\frac{x y}{z r} + \frac{y z}{x r}}{1 - \frac{x y}{z r} \frac{y z}{x r}} ⎞ ⎟ ⎟ ⎠ + {tan}^{- 1} (\frac{x z}{y r})$
$= {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{y (x^{2} + z^{2})}{x z r}}{\frac{r^{2} - y^{2}}{r^{2}}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + {tan}^{- 1} (\frac{x z}{y r}) (∵ x^{2} + y^{2} + z^{2} = r^{2})$
$= {tan}^{-} 1 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{y r (x^{2} + z^{2})}{x z}}{x^{2} + z^{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ + {tan}^{- 1} (\frac{x z}{y r})$
$= {tan}^{- 1} (\frac{y r}{x z}) + {tan}^{- 1} (\frac{x z}{y r})$
$= \frac{π}{2}$

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