Question

The sum of ${\cot }^{-1}3+{\cot }^{-1}7+{\cot }^{-1}13+{\cot }^{-1}21+{\cot }^{-1}31$ is

• A. ${\cot }^{-1}\frac{7}{5}$
• B. ${\cot }^{-1}\frac{5}{7}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A ${\cot }^{-1}\frac{7}{5}$

$=\left(ta{n}^{-1}\right(\frac{1}{3})+ta{n}^{-1}(\frac{1}{7}\left)\right)+\left(ta{n}^{-1}\right(\frac{1}{13})+ta{n}^{-1}(\frac{1}{21}\left)\right)+ta{n}^{-1}\left(\frac{1}{31}\right)$
[$ta{n}^{-1}\left(a\right)+ta{n}^{-1}\left(b\right)$ $=ta{n}^{-1}\left(\frac{a+b}{1-ab}\right)$]
$=ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)+ta{n}^{-1}\left(\frac{1}{31}\right)$
$=ta{n}^{-1}\left(\frac{2}{3}\right)+ta{n}^{-1}\left(\frac{1}{31}\right)$
$=ta{n}^{-1}\left(\frac{5}{7}\right)$
$=co{t}^{-1}\left(\frac{7}{5}\right)$