Question

$2\cot \left({\cot }^{-1}\right(3)+{\cot }^{-1}(7)+{\cot }^{-1}(13)+{\cot }^{-1}(21\left)\right)$ has the value equal to

• A. $\frac{1}{2}$
• B. $2$
• C. $3$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $3$

Consider $\sum {\cot }^{-1}(n^2+n+1)$ where $n=1,2,3,4$
Since ${\tan }^{-1}x={\cot }^{-1}\frac{1}{x}$ for $x>0,$ we get
$\sum {\tan }^{-1}\frac{1}{1+n(n+1)}$
$=\sum {\tan }^{-1}\frac{(n+1)-n}{1+n(n+1)}$
$=\sum \left({\tan }^{-1}\right(n+1)-{\tan }^{-1}(n\left)\right)$

$T_n={\tan }^{-1}(n+1)-{\tan }^{-1}\left(n\right)T_1={\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)T_2={\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)T_3={\tan }^{-1}\left(4\right)-{\tan }^{-1}\left(3\right)T_4={\tan }^{-1}\left(5\right)-{\tan }^{-1}\left(4\right)$
$\therefore S=T_1+T_2+T_3+T_4$
$={\tan }^{-1}\left(5\right)-{\tan }^{-1}\left(1\right)$
$={\tan }^{-1}\frac{5-1}{1+5}={\tan }^{-1}\frac{2}{3}$
$\therefore 2\cot \left({\cot }^{-1}\frac{3}{2}\right)=2\cdot \frac{3}{2}=3$