Question

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

Answer 1

Hint: Assume three variables (say x, y, z) for ${{\tan }^{-1}}\left( 1 \right),{{\tan }^{-1}}2\ and {{\tan }^{-1}}3$ respectively. Then use the formula,
$''\tan \left( x+y+z \right)=\dfrac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan z.\tan y-\tan y\tan z}''$.
Put the value of $\tan x,\tan y\ and\ \tan z$ to get the value of $\tan \left( x+y+z \right)$ and then with the help of $\tan \left( x+y+z \right)$, find the value of $x+y+z$.

Complete Step-by-step answer:
To prove: ${{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right)=\pi $
Proof:
$\begin{align}
  & LHS={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right) \\
 & RHS=\pi \\
\end{align}$
We have to prove LHS = RHS
Let us start with LHS.
$LHS={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right)$
Let us assume $x={{\tan }^{-1}}1$
Taking tan both sides, we will get,
$\begin{align}
  & \tan \left( x \right)=\tan \left( {{\tan }^{-1}}1 \right) \\
 & \Rightarrow \tan x=1 \\
\end{align}$
Similarly, let us assume $y={{\tan }^{-1}}2$
Taking tan both sides, we will get,
$\begin{align}
  & \tan y=\tan \left( {{\tan }^{-1}}2 \right) \\
 & \Rightarrow \tan y=2 \\
\end{align}$
Similarly, let us assume $z={{\tan }^{-1}}3$
Taking tan both sides, we will get,
$\begin{align}
  & \tan z=\tan \left( {{\tan }^{-1}}3 \right) \\
 & \Rightarrow \tan z=3 \\
\end{align}$
Now, let us use the formula,
$\tan \left( x+y+z \right)=\dfrac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan z.\tan y-\tan y\tan z}$
On putting tan x = 1, tan y = 2 and tan z = 3 as calculated above, we will get,

$\begin{align}
  & \Rightarrow \tan \left( x+y+z \right)=\dfrac{1+2+3-\left( 1 \right)\left( 2 \right)\left( 3 \right)}{1-\left( 1 \right)\left( 2 \right)-\left( 1 \right)\left( 3 \right)-\left( 2 \right)\left( 3 \right)} \\
 & \Rightarrow \tan \left( x+y+z \right)=\dfrac{6-6}{1-2-3-6} \\
 & \Rightarrow \tan \left( x+y+z \right)=\dfrac{0}{-10} \\
 & \Rightarrow \tan \left( x+y+z \right)=0 \\
\end{align}$
We know,
 $\begin{align}
  & \tan \pi =0 \\
 & \Rightarrow x+y+z=\pi \\
\end{align}$
Replace $x\ with\ {{\tan }^{-1}}1\ and\ y\ with\ {{\tan }^{-1}}2\ and\ z\ with\ {{\tan }^{-1}}3$, we will get,
${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3=\pi $
Proved.

Note: Note that tan (0) is also equal to 0. But ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$ can’t be equal to zero. As tan of these three angles are positive which mean all these angles are greater than zero and thus their sum can’t be zero. Students can make mistakes in the last step by taking x+y+z=0.