First slide

Properties of ITF

Question

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

Difficult

A

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

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

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

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

Solution

$W e h a v e T a n [T a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + T a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + . . . . . . + T a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] = T a n [T a n^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + T a n^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} a_{3}}) + . . . . . . + T a n^{- 1} (\frac{a_{n} - a_{n - 1}}{1 + a_{n - 1} a_{n}})] (∵ d = a_{2} - a_{1} = a_{3} - a_{2} = . . . . . . = a_{n} - a_{n - 1}) = T a n [T a n^{- 1} (a_{2}) - T a n^{- 1} (a_{1}) + T a n^{- 1} (a_{3}) - T a n^{- 1} (a_{2}) + . . . . . . + T a n^{- 1} (a_{n}) - T a n^{- 1} (a_{n - 1})] = T a n [T a n^{- 1} (a_{n}) - T a n^{- 1} (a_{1})] = T a n [T a n^{- 1} (\frac{a_{n} - a_{1}}{1 + a_{n} a_{1}})] = \frac{(n - 1) d}{1 + a_{1} a_{n}} (∵ a_{n} = a_{1} + (n - 1) d)$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App