If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in arithmetic progression with common difference $d,$ then evaluate the following expression :
$tan {{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + {tan}^{- 1} (\frac{d}{1 + a_{3} a_{4}}) + . . . . . . + {tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})}$

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Solution

$d = a_{2} - a_{1} = a_{3} - a_{2} . . . . = a_{n} - a_{n - 1}$
$tan {{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} \frac{d}{1 + a_{2} a_{3}} + . . . . {tan}^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}}$
$= tan {{tan}^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + {tan}^{- 1} \frac{a_{3} - a_{2}}{1 + a_{2} a_{3}} + . . . . . . {tan}^{- 1} \frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}}}$
$= tan {a_{2} - a_{1} + a_{3} - a_{2} + a_{4} - a_{3} + . . . . . . a_{n} - a_{n - 1}}$
$= tan {a_{n} - a_{1}}$
$= tan {a_{1} + (n - 1) d - a_{1}}$
$= tan {(n - 1) d}$

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If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ is an arithmetic progression with common difference d, then evaluate the following expression.

$t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + t a n^{- 1} (\frac{d}{1 + a_{3} a_{4}}) + \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$

Q.
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a_{1}, a_{2}, a_{3} \dots \dots . . a_{n}

are in A.P. with common difference d, then

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}}) + \dots \dots \dots \dots \dots \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

Q. If

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

, where

a_{1}, a_{2}, a_{3}, \dots a_{n}

are in A.P. with common difference d, then

t a n θ =

Q. If

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

, where

a_{1}, a_{2}, a_{3}, \dots a_{n}

are in A.P. with common difference d, then

t a n θ =

Q. If

a_{1}, a_{2}, a_{3}, . . ., a_{n}

are in A.P. with common difference

d

where

a_{r} > 0, \forall r = 1, 2, 3, . . ., n,

then

tan [{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots + {tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

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If a1,a2,a3,…,an are in arithmetic progression with common difference d, then evaluate the following expression :tan{tan−1(d1+a1a2)+tan−1(d1+a2a3)+tan−1(d1+a3a4)+......+tan−1(d1+an−1an)}

d=a2−a1=a3−a2....=an−an−1tan{tan−1(d1+a1a2)+tan−1d1+a2a3+....tan−1d1+an−1an}=tan{tan−1(a2−a11+a1a2)+tan−1a3−a21+a2a3+......tan−1an−an−11+an−1an}=tan{a2−a1+a3−a2+a4−a3+......an−an−1}=tan{an−a1}=tan{a1+(n−1)d−a1}=tan{(n−1)d}