Question

$ta{n}^{-1}1+ta{n}^{-1}2+ta{n}^{-1}3$ is equal to

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $2\pi$
• D. $\frac{26\pi }{24}$

Answer 1

The correct option is B $\pi$

Given ${\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)$

$=\pi +{\tan }^{-1}\left(\frac{3}{1-2}\right)+{\tan }^{-1}\left(3\right)$ since ${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)=\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$ as $xy>1$

$=\pi +{\tan }^{-1}(-3)+{\tan }^{-1}\left(3\right)$

$=\pi -{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(3\right)$

$=\pi$