First slide

Properties of ITF

Question

If $\tan^{- 1} \frac{1}{1 + 2} + \tan^{- 1} \frac{1}{1 + 2 .3} + \tan^{- 1} \frac{1}{1 + 3 .4} + \dots + \tan^{- 1} \frac{1}{1 + n (n + 1)} = \tan^{- 1} x$ , then x is equal to

Moderate

A

$\frac{n}{n + 2}$
B

$\frac{n}{n + 1}$
C

$\frac{n - 1}{n + 2}$
D

none of these

Solution

We have,
$\begin{array}{rcl} L.H.S. & = & \tan^{- 1} \frac{2 - 1}{1 + 2 .1} + \tan^{- 1} \frac{3 - 2}{1 + 3 .2} + \tan^{- 1} \frac{4 - 3}{1 + 4 .3} +, \dots, + \tan^{- 1} \frac{\bar{n + 1} - n}{1 + (n + 1) n} \\ = & (\tan^{- 1} 2 - \tan^{- 1} 1) + (\tan^{- 1} 3 - \tan^{- 1} 2) + (\tan^{- 1} 4 - \tan^{- 1} 3) +, \dots, + (\tan^{- 1} (n + 1) - \tan^{- 1} n) \\ = & \tan^{- 1} (n + 1) - \tan^{- 1} 1 = \tan^{- 1} (\frac{(n + 1) - 1}{1 + (n + 1) 1}) \\ = & \tan^{- 1} \frac{n}{n + 2} = \tan^{- 1} x (given) \end{array}$
$∴ x = \frac{n}{n + 2}$

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