Question

The value of ${\cot }^{-1}\left(\frac{1}{3}\right)+{\cot }^{-1}\left(\frac{1}{2}\right)+{\cot }^{-1}\left(1\right)$ is

• A. $\frac{3\pi }{2}$
• B. $0$
• C. $\pi$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is C $\pi$

$S={\cot }^{-1}\left(\frac{1}{3}\right)+{\cot }^{-1}\left(\frac{1}{2}\right)+{\cot }^{-1}\left(1\right)$
$={\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(1\right)$

$\because {\tan }^{-1}x+{\tan }^{-1}y=\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right),x>0,y>0,xy>1$
$\therefore S={\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(1\right)$
$=\pi +{\tan }^{-1}\left(\frac{3+2}{1-3\times 2}\right)+\frac{\pi }{4}$
$=\pi +{\tan }^{-1}(-1)+\frac{\pi }{4}=\pi -\frac{\pi }{4}+\frac{\pi }{4}=\pi$