Question

Sum of series ${\cot }^{-1}\left(\frac{5}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{9}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{15}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{23}{\sqrt{3}}\right)+....\infty =$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is B $\frac{\pi }{3}$

${\tan }^{-1}\left(\frac{\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}}{1+\frac{2}{\sqrt{3}}.\frac{1}{\sqrt{3}}}\right)+{\tan }^{-1}\left(\frac{\frac{3}{\sqrt{3}}-\frac{2}{\sqrt{3}}}{1+\frac{3}{\sqrt{3}}.\frac{2}{\sqrt{3}}}\right)+....+{\tan }^{-1}\left(\frac{\frac{n+1}{\sqrt{3}}-\frac{n}{\sqrt{3}}}{1+\frac{n+1}{\sqrt{3}}.\frac{n}{\sqrt{3}}}\right)+...\infty$

$=\underset{n\to \infty }{lim}\left\{\right({\tan }^{-1}\frac{2}{\sqrt{3}}-{\tan }^{-1}\frac{1}{\sqrt{3}})+({\tan }^{-1}\frac{3}{\sqrt{3}}-{\tan }^{-1}\frac{2}{\sqrt{3}})+...+({\tan }^{-1}\left(\frac{n+1}{\sqrt{3}}\right)-{\tan }^{-1}\left)\right(\frac{n}{\sqrt{3}}\left)\right\}$

$=\underset{n\to \infty }{lim}\left\{{\tan }^{-1}\right(\frac{n+1}{\sqrt{3}})-{\tan }^{-1}(\frac{1}{\sqrt{3}}\left)\right\}=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$