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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}

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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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