${tan}^{- 1} a + {tan}^{- 1} b,$ where $a > 0, b > 0, a b > 1,$ is equal to

{tan}^{- 1} \frac{a + b}{1 - a b}

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{tan}^{- 1} \frac{a + b}{1 - a b} - π

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π + {tan}^{- 1} \frac{a + b}{1 - a b}

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none of these

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Solution

The correct option is D $π + {tan}^{- 1} \frac{a + b}{1 - a b}$
Given, ${tan}^{- 1} (a) + {tan}^{- 1} (b)$
$= {tan}^{- 1} (\frac{a + b}{1 - a b})$
However, $a b > 1$
Therefore,
$π + {tan}^{- 1} (\frac{a + b}{1 - a b})$

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