If $a_{1}, a_{2}, a_{3}, \dots . a_{n}$ in A.P. with common difference ‘d’, then $\tan [\tan^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {Tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + \dots . . + {Tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =$

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$\frac{(n - 1) d}{1 + a_{1} a_{n}}$

$\frac{a_{n} - a_{1}}{a_{n} + a_{1}}$

$\frac{nd}{1 + a_{1} a_{n}}$

$\frac{(n - 1) d}{a_{1} + a_{n}}$

answer is B.

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Detailed Solution

$\tan [\tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{3} a_{2}}) + \dots . + \tan^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n} a_{n - 1}})]$ $\begin{array}{l} = \tan [(\tan^{- 1} a_{2} - \tan^{- 1} a_{1}) + (\tan^{- 1} a_{3} - \tan^{- 1} a_{2}) + . . . + (\tan^{- 1} a_{n} - \tan^{- 1} a_{n - 1})] = \tan [\tan^{- 1} a_{n} - \tan^{- 1} a_{1}] = \frac{a_{n} - a_{1}}{1 + a_{1} a_{n}} \\ = \frac{a_{1} + (n - 1) d - a_{1}}{1 + a_{1} a_{n}} = \frac{(n - 1) d}{1 + a_{1} a_{n}} \end{array}$