The value of cot(19∑n=1cot⁻¹(1+n∑p=12p)) is:

The correct option is C
2119
cot(19∑n=1cot⁻¹(1+n(n+1)))
cot(19∑n=1cot⁻¹(n²+n+1))=cot(19∑n=1tan⁻¹1/(1+n(n+1)))
19∑n=1(tan⁻¹(n+1)-tan⁻¹n)
cot(tan⁻¹20-tan⁻¹1)=cotA
cotB+1
cotB-cotA
(Where tanA=20, tanB=1)
1(1/20)+1
1-1/20=21/19