Q

# Help me answer: - The value of is: - Trigonometry - JEE Main

The value of $\cot \left(\sum_{n=1}^{19}\cot^{-1}\left(1 + \sum_{p=1}^{n}2p \right )\right )$ is:

• Option 1)

$\frac{19}{21}$

• Option 2)

$\frac{23}{22}$

• Option 3)

$\frac{22}{23}$

• Option 4)

$\frac{21}{19}$

Views

Important Results of Inverse Trigonometric Functions -

$\cot ^{-1}\left ( \cot \Theta \right )= \Theta$

- wherein

if $0< \Theta < \pi$

Results of Compound Angles -

$\cot \left ( A-B \right )= \frac{\cot A\cot B+1}{\cot B-\cot A}$

- wherein

Where A and B are two angles.

$\cot \left(\sum_{n=1}^{19}\cot^{-1}\left(1 + \sum_{p=1}^{n}2p \right )\right )$

From concept learnt $\sum n=\frac{n(n+1)}{2}$

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

$=\cot [\sum_{n=1}^{19}\cot^{-1}(n^{2}+n+1)]$

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

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

$=\cot [\sum_{n=1}^{19}\tan^{-1}(20)-\tan^{-1}1]$

$=\frac{\cot A\cot B+1}{\cot B-\cot A}$

When $\tan A=20 \: \: and\: \: \tan B=1$

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

$=\frac{21}{19}$

Option 1)

$\frac{19}{21}$

Option 2)

$\frac{23}{22}$

Option 3)

$\frac{22}{23}$

Option 4)

$\frac{21}{19}$

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