First slide

Properties of ITF

Question

The value of $\cot [\sum_{n = 1}^{19} \cot^{- 1} \{1 + \sum_{p = 1}^{n} 2 p\}]$ is

Moderate

A

$\frac{22}{23}$
B

$\frac{23}{22}$
C

$\frac{21}{19}$
D

$\frac{19}{21}$

Solution

$1 + \sum_{p = 1}^{n} 2 p = 1 + n (n + 1) = n^{2} + n + 1$
$∴ \cot^{- 1} \{1 + \sum_{p = 1}^{n} 2 p\} = \cot^{- 1} (n^{2} + n + 1) = \tan^{- 1} \frac{1}{1 + n (n + 1)}$
$\begin{aligned} = \tan^{- 1} \{\frac{(n + 1) - n}{1 + n (n + 1)}\} \\ = \tan^{- 1} (n + 1) - \tan^{1} n \end{aligned}$
$\begin{aligned} \Rightarrow \sum_{n = 1}^{19} \cot^{- 1} \{1 + \sum_{p = 1}^{n} 2 p\} = \sum_{n = 1}^{19} \{\tan^{- 1} (n + 1) - \tan^{- 1} n\} \\ = \tan^{- 1} 20 - \tan^{- 1} 1 = \tan^{- 1} \frac{19}{21} \end{aligned}$
$\begin{aligned} \Rightarrow \cot [\sum_{n = 1}^{19} \cot^{- 1} \{1 + \sum_{p = 1}^{n} 2 p\}] = \cot (\tan^{- 1} \frac{19}{21}) \\ = \cot (\cot^{- 1} \frac{21}{19}) = \frac{21}{19} \end{aligned}$

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