If a1-a2-a3- are in A-P- with common difference -d-then tantan - 1 d-1 - a1a2 - tan - 1 d-1 - a2a3 - - - d-1 - an - 1an is equal to

The correct option is B ( n 1)d/1 + a_1a_n a_1, a_2, ... , a_n are in A.P with common difference d. a_2 = a_1 + d a_3 = a_1 + 2d a_n ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If a1,a2,a3...

Question

If $a_{1,} a_{2}, a_{3} \dots$ are in A.P. with common difference 'd',then $tan {{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots + (\frac{d}{1 + a_{n - 1} a_{n}})}$ is equal to

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

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

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\frac{(n - 1) d}{1 + a_{2} a_{n}}

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

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Solution

The correct option is B $\frac{(n - 1) d}{1 + a_{1} a_{n}}$
$a_{1}, a_{2}, . . ., a_{n}$ are in A.P with common difference d.
$a_{2} = a_{1} + d$
$a_{3} = a_{1} + 2 d$
$a_{n} = a_{1} + (n - 1) d$
$t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) = t a n^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) = t a n^{- 1} (a_{2}) - t a n^{- 1} (a_{1})$
$t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) = t a n^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} a_{3}}) = t a n^{- 1} (a_{3}) - t a n^{- 1} (a_{2})$
$t a n^{- 1} (\frac{d}{1 + a_{n} + a_{n}}) = t a n^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}}) = t a n^{- 1} (a_{n}) - t a n^{- 1} (a_{n - 1})$
$∴ t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + . . + t a n^{- 1} (\frac{d}{1 + a_{n} + a_{n}})$
$= t a n^{- 1} (a_{n}) - t a n^{- 1} (a_{1})$
$= t a n^{- 1} (\frac{a_{n} - a_{1}}{1 + a_{1} a_{n}})$
$= t a n^{- 1} (\frac{a_{1} + (n - 1) d - a_{1}}{1 + a_{1} a_{n}})$
$= t a n^{- 1} (\frac{(n - 1) d}{1 + a_{1} a_{n}})$
$∴ t a n {t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + . . . + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})}$
$= t a n {t a n^{- 1} (\frac{(n - 1) d}{1 + a_{1} a_{n}})}$
$= \frac{(n - 1) d}{1 + a_{1} a_{n}}$

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If a1,a2,a3… are in A.P. with common difference 'd',then tan{tan−1(d1+a1a2)+tan−1(d1+a2a3)+…+(d1+an−1an)} is equal to

If $a_{1,} a_{2}, a_{3} \dots$ are in A.P. with common difference 'd',then $tan {{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots + (\frac{d}{1 + a_{n - 1} a_{n}})}$ is equal to