Question

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

Answer 1

Proof :

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

Taking L.H.S ${\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(2\right)+{\tan }^{-1}\left(3\right)$

Let ${\tan }^{-1}1=x\Rightarrow \tan x=1,{\tan }^{-1}2=y\Rightarrow \tan y=2,{\tan }^{-1}z=3\Rightarrow \tan 3=z$

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

$$\begin{gathered} \Rightarrow \tan (x+y+z)=\frac{1+2+3-1.2.3}{1-1.2-1.3-2.3} \\ \Rightarrow \tan (x+y+z)=0 \\ \Rightarrow \tan (x+y+z)=\tan \pi \\ \Rightarrow (x+y+y)=\pi \end{gathered}$$

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

Hence proved.