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The value of $\tan^{- 1} (1) + \tan^{- 1} (0) + \tan^{- 1} (2) + \tan^{- 1} (3)$ is equal to

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$π$

None of these

$\frac{5 π}{4}$

$\frac{π}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} \tan^{- 1} (1) + \tan^{- 1} (0) + \tan^{- 1} (2) + \tan^{- 1} (3) \\ = \frac{π}{4} + π + \tan^{- 1} (\frac{2 + 3}{1 - 2.3}) (as 2.3 > 1) \\ = \frac{5 π}{4} + \tan^{- 1} (- 1) = \frac{5 π}{4} - \frac{π}{4} = π \end{array}$

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