The value of

Question:

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

 

  1. (1) $\frac{21}{19}$

  2. (2) $\frac{19}{21}$

  3. (3) $\frac{22}{23}$

  4. (4) $\frac{23}{22}$


Correct Option: 1

Solution:

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

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

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

$\left[\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right):\right.$ for $\left.x>0\right]$

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

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

$=\cot \left(\tan ^{-1}\left(\frac{20-1}{1+20 \times 1}\right)\right)$

$=\cot \left(\tan ^{-1}\left(\frac{19}{21}\right)\right)=\cot \cot ^{-1}\left(\frac{21}{19}\right)=\frac{21}{19}$

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