Question

Prove that ${\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)=\pi$

Answer 1

Let ${\tan }^{-1}\left(1\right)=x$

$\Rightarrow 1=\tan x$

Let ${\tan }^{-1}\left(2\right)=y$

$\Rightarrow 2=\tan y$

Let ${\tan }^{-1}\left(3\right)=z$

$\Rightarrow 3=\tan z$

$\tan (x+y+z)=\frac{\tan x+\tan y+\tan z-\tan x.\tan y.\tan z}{1-\tan x.\tan y-\tan x.\tan z-\tan y.\tan z}$

$=\frac{1+2+3-1\times 2\times 3}{1-1\times 2-1\times 3-2\times 3}$

$=0$

$\Rightarrow x+y+z=\pi$

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

( $x+y+z$ cannot be equal to zero, because

${\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)$ will have some value greater than zero )