Question

The value of $5\cot \left(\sum _{k=1}^5\left({\cot }^{-1}\left(k^2+k+1\right)\right)\right)$ is equal to

• A. 6
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (C)

$\text{ Consider }\sum _{k=1}^5\left({\tan }^{-1}\left(\frac{(k+1)-k}{1+k(k+1)}\right)\right))={\tan }^{-1}(k+1)-{\tan }^{-1}k$
$T_1={\tan }^{-1}(2)-{\tan }^{-1}(1);T_2={\tan }^{-1}(3)-{\tan }^{-1}(2)\text{ and so on }$
$\text{ hence }S={\tan }^{-1}(6)-{\tan }^{-1}(1)={\tan }^{-1}\left(\frac{5}{7}\right)={\cot }^{-1}\left(\frac{7}{5}\right)$
$\therefore 5\cot \left({\cot }^{-1}\frac{7}{5}\right)=7$