Question

Prove that $\tan \left( \arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n} \right)=1$

Answer 1

Hint: Put $\arctan \dfrac{m}{n}=x,\arctan \dfrac{m-n}{m+n}=y$ and use the identity $\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$. Use $\tan \left( \arctan x \right)=x$. Simplify to get the result. Alternatively, you can use the identity $\arctan x-\arctan y=n\pi +\arctan \dfrac{x-y}{1+xy},n\in \mathbb{Z}$, where n is suitably chosen integer. “n” is chosen so that the sum $n\pi +\arctan \dfrac{x-y}{1+xy}$ is within the interval $\left( -\pi ,\pi \right)$.



Complete step-by-step answer:
Let $\arctan \dfrac{m}{n}=x$ and $\arctan \dfrac{m-n}{m+n}=y$.
Hence we have LHS = tan(x-y).
We know that $\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$
Using the above identity, we get
LHS $=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$.
Reverting to original variables, we get
LHS $=\dfrac{\tan \left( \arctan \dfrac{m}{n} \right)-\tan \left( \arctan \dfrac{m-n}{m+n} \right)}{1+\tan \left( \arctan \dfrac{m}{n} \right)\tan \left( \arctan \dfrac{m-n}{m+n} \right)}$
We know that $\tan \left( \arctan x \right)=x$
Using the above identity, we get
LHS $=\dfrac{\dfrac{m}{n}-\dfrac{m-n}{m+n}}{1+\dfrac{m}{n}\dfrac{m-n}{m+n}}$
Multiplying numerator and denominator by n(m+n), we get
LHS
 $\begin{align}
  & =\dfrac{\dfrac{m}{n}n\left( m+n \right)-\dfrac{m-n}{m+n}n\left( m+n \right)}{n\left( m+n \right)+\dfrac{m}{n}\dfrac{m-n}{m+n}n\left( m+n \right)} \\
 & =\dfrac{m\left( m+n \right)-n\left( m-n \right)}{n\left( m+n \right)+m\left( m-n \right)} \\
\end{align}$
We know that a(b+c) = ab+ac and a(b-c) = ab-ac {Distributive law of multiplication over addition}
Using the above property, we get
LHS
$\begin{align}
  & =\dfrac{{{m}^{2}}+mn-nm+{{n}^{2}}}{mn+{{n}^{2}}+{{m}^{2}}-mn} \\
 & =\dfrac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}}=1 \\
\end{align}$
Hence LHS = RHS.
Hence, we have $\tan \left( \arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n} \right)=1$
Hence proved.
Note: Alternate Solution:
We know that $\arctan x-\arctan y=n\pi +\arctan \dfrac{x-y}{1+xy},n\in \mathbb{Z}$, where n is suitably chosen so that $n\pi +\arctan \dfrac{x-y}{1+xy}$ is within the interval $\left( -\pi ,\pi \right)$
Put $x=\dfrac{m}{n}$ , and $y=\dfrac{m-n}{m+n}$ in the above identity, we get
$\arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n}=k\pi +\arctan \left( \dfrac{\dfrac{m}{n}-\dfrac{m-n}{m+n}}{1+\dfrac{m}{n}\dfrac{m-n}{m+n}} \right)$, for some suitably chosen k.
Simplifying the argument of arctan we get

$\arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n}=k\pi +\arctan \left( \dfrac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}} \right)=k\pi +\dfrac{\pi }{4}$
Since $\arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n}\in \left( -\pi ,\pi \right)$we have
$k\pi +\dfrac{\pi }{4}\in \left( -\pi ,\pi \right),k\in \mathbb{Z}$
Hence we have k = 0.
Hence $\arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n}=\dfrac{\pi }{4}$
Hence we have
$\tan \left( \arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n} \right)=\tan \dfrac{\pi }{4}=1$
Hence proved.
[2] Observe that in the alternative solution, we need not find the value of k since $\tan \left( n\pi +x \right)=\tan x,n\in \mathbb{Z}$.
Hence if we take tangents on both sides, we get
$\tan \left( \arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n} \right)=\tan \left( k\pi +\dfrac{\pi }{4} \right)=\tan \dfrac{\pi }{4}=1$.
[3]. Since tanx is not a one-one function, its inverse is defined only within some intervals. The principal interval for arctanx is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. Whenever not explicitly mentioned, we take arctanx in the principal interval. Hence we have
$\begin{align}
  & \arctan x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right),\arctan y\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) \\
 & \Rightarrow \arctan x-\arctan y\in \left( -\pi ,\pi \right) \\
\end{align}$
This is why we chose the value of n, so that $n\pi +\arctan \dfrac{x-y}{1+xy}$ is within the interval $\left( -\pi ,\pi \right)$. It can also be proved that unique n satisfies this property.