SATHEE: Inverse Circular Functions 2 Question 2

Inverse Circular Functions 2 Question 2

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

(a) $\frac{23}{22}$

(b) $\frac{21}{19}$

(d) $\frac{22}{23}$

(2019 Main, 10 Jan II)

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Answer:

Correct Answer: 2. (b)

Solution:

Consider, $\cot \sum_{n = 1}^{19} \cot^{- 1} 1 + \sum_{p = 1}^{n} 2 p$

$\begin{aligned} = \cot \sum_{n = 1}^{19} \cot^{- 1} (1 + n (n + 1)) ∵ \sum_{n = 1}^{19} p = \frac{n (n + 1)}{2} \\ = \cot \sum_{n = 1}^{19} \cot^{- 1} (1 + n + n^{2}) \\ = \cot \sum_{n = 1}^{19} \tan^{- 1} \frac{1}{1 + n (n + 1)} \end{aligned}$

$[∵ \cot^{- 1} x = \tan^{- 1} \frac{1}{x}, if x > 0]$

$= \cot \sum_{n = 1}^{19} \tan^{- 1} \frac{(n + 1) - n}{1 + n (n + 1)} [$ put $1 = (n + 1) - n]$

$= \cot \sum_{n = 1}^{19} (\tan^{- 1} (n + 1) - \tan^{- 1} n)$

$∵ \tan^{- 1} \frac{x - y}{1 + x y} = \tan^{- 1} x - \tan^{- 1} y$

$= \cot [(\tan^{- 1} 2 - \tan^{- 1} 1) + (\tan^{- 1} 3 - \tan^{- 1} 2) +$

$= \cot (\tan^{- 1} 20 - \tan^{- 1} 1)$ .. $+ (\tan^{- 1} 20 - \tan^{- 1} 19)]$

$= \cot \frac{π}{2} - \cot^{- 1} 20 - \frac{π}{2} - \cot^{- 1} 1$

$[∵ \tan^{- 1} x + \cot^{- 1} x = π / 2]$

$= \cot (\cot^{- 1} 1 - \cot^{- 1} 20)$

$= \frac{\cot (\cot^{- 1} 1) \cot (\cot^{- 1} 20) + 1}{\cot (\cot^{- 1} 20) - \cot (\cot^{- 1} 1)}$

$= \frac{(1 \times 20) + 1}{20 - 1}$

$[∵ \cot (A - B) = \frac{\cot A \cot B + 1}{\cot B - \cot A}$

$= \frac{21}{19}$ $[∵ \cot (\cot^{- 1} x) = x]$

Inverse Circular Functions 2 Question 2

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

Answer:

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Inverse Circular Functions 2 Question 2

2. The value of cot ∑n=119cot−1⁡1+∑p=1n2p is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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