If a1,a2,a3…….an are in A.P. with common difference ‘d’, then tan⁡Tan−1⁡d1+a1a2+Tan−1⁡d1+a2a3+……+Tan−1⁡d1+an−1an=

tan⁡tan−1⁡a2−a11+a1a2+tan−1⁡a3−a21+a3a2+……+tan−1⁡an−an−11+anan−1=a1+(n−1)d−a11+a1an

If a1,a2,a3…….an are in A.P. with common difference ‘d’, then tan⁡Tan−1⁡d1+a1a2+Tan−1⁡d1+a2a3+……+Tan−1⁡d1+an−1an=

tan⁡tan−1⁡a2−a11+a1a2+tan−1⁡a3−a21+a3a2+……+tan−1⁡an−an−11+anan−1=a1+(n−1)d−a11+a1an

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

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

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

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

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

answer is B.

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$\tan [\tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{3} a_{2}}) + \dots \dots +$

$\tan^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n} a_{n - 1}})]$

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

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If a1,a2,a3…….an are in A.P. with common difference ‘d’, then tan⁡Tan−1⁡d1+a1a2+Tan−1⁡d1+a2a3+……+Tan−1⁡d1+an−1an=

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