The value of (_n=1^19^-1(1+_p=1^n2p)) is:

Question

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

Accepted Answer

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

$$\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)$$

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

$$\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta  + 1} \over {\cot \beta  - \cot A}}$$

(Where tanA $=$ 20, tanB $=$ 1)   $${{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}$$

Solutions · 1