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Mathematics Question on Properties of Inverse Trigonometric Functions

The value of cot⁡(∑n=119cot⁡−1(1+∑p=1n2p))\cot\left(\sum\limits^{19}_{n=1} \cot^{-1} \left(1+ \sum\limits^{n}_{p=1} 2p\right)\right) is :

A

2223\frac{22}{23}

B

2322\frac{23}{22}

C

2119\frac{21}{19}

D

1921\frac{19}{21}

Answer

2119\frac{21}{19}

Explanation

Solution

cot⁡(∑n=119cot⁡−1(1+n(n+1)))\cot\left(\sum^{19}_{n=1} \cot^{-1} \left(1+n\left(n+1\right)\right)\right)
cot⁡(∑n=119cot⁡−1(n2+n+1))=cot⁡(∑n=119tan⁡−111+n(n+1))\cot\left(\sum^{19}_{n=1} \cot^{-1}\left(n^{2}+n+1\right)\right) = \cot\left(\sum^{19}_{n=1} \tan^{-1} \frac{1}{1+n\left(n+1\right)}\right)
∑n=119(tan⁡−1(n+1)−tan⁡−1n)\sum^{19}_{n=1} \left(\tan^{-1}\left(n+1\right)-\tan^{-1}n\right)
cot⁡(tan⁡−120−tan⁡−11)=cot⁡Acot⁡β+1cot⁡β−cot⁡A\cot\left(\tan^{-1}20 -\tan^{-1}1\right)= \frac{\cot A \cot\beta+1}{\cot\beta-\cot A}
(Where tan⁡A=20,tan⁡B=1\tan A=20, \tan B=1 )
1(120)+11−120=2119\frac{1\left(\frac{1}{20}\right)+1}{1- \frac{1}{20}} = \frac{21}{19}