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

Prove that tan⁡(arctan⁡mn−arctan⁡m−nm+n)=1\tan \left( \arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n} \right)=1

Explanation

Solution

Hint: Put arctan⁡mn=x,arctan⁡m−nm+n=y\arctan \dfrac{m}{n}=x,\arctan \dfrac{m-n}{m+n}=y and use the identity tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}. Use tan⁡(arctan⁡x)=x\tan \left( \arctan x \right)=x. Simplify to get the result. Alternatively, you can use the identity arctan⁡x−arctan⁡y=nπ+arctan⁡x−y1+xy,n∈Z\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 nπ+arctan⁡x−y1+xyn\pi +\arctan \dfrac{x-y}{1+xy} is within the interval (−π,π)\left( -\pi ,\pi \right).

Complete step-by-step answer:
Let arctan⁡mn=x\arctan \dfrac{m}{n}=x and arctan⁡m−nm+n=y\arctan \dfrac{m-n}{m+n}=y.
Hence we have LHS = tan(x-y).
We know that tan⁡(x−y)=tan⁡x−tan⁡y1+tan⁡xtan⁡y\tan \left( x-y \right)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}
Using the above identity, we get
LHS =tan⁡x−tan⁡y1+tan⁡xtan⁡y=\dfrac{\tan x-\tan y}{1+\tan x\tan y}.
Reverting to original variables, we get
LHS =tan⁡(arctan⁡mn)−tan⁡(arctan⁡m−nm+n)1+tan⁡(arctan⁡mn)tan⁡(arctan⁡m−nm+n)=\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⁡(arctan⁡x)=x\tan \left( \arctan x \right)=x
Using the above identity, we get
LHS =mn−m−nm+n1+mnm−nm+n=\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
=mnn(m+n)−m−nm+nn(m+n)n(m+n)+mnm−nm+nn(m+n) =m(m+n)−n(m−n)n(m+n)+m(m−n) \begin{aligned} & =\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{aligned}
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
=m2+mn−nm+n2mn+n2+m2−mn =m2+n2m2+n2=1 \begin{aligned} & =\dfrac{{{m}^{2}}+mn-nm+{{n}^{2}}}{mn+{{n}^{2}}+{{m}^{2}}-mn} \\\ & =\dfrac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}}=1 \\\ \end{aligned}
Hence LHS = RHS.
Hence, we have tan⁡(arctan⁡mn−arctan⁡m−nm+n)=1\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=nπ+arctan⁡x−y1+xy,n∈Z\arctan x-\arctan y=n\pi +\arctan \dfrac{x-y}{1+xy},n\in \mathbb{Z}, where n is suitably chosen so that nπ+arctan⁡x−y1+xyn\pi +\arctan \dfrac{x-y}{1+xy} is within the interval (−π,π)\left( -\pi ,\pi \right)
Put x=mnx=\dfrac{m}{n} , and y=m−nm+ny=\dfrac{m-n}{m+n} in the above identity, we get
arctan⁡mn−arctan⁡m−nm+n=kπ+arctan⁡(mn−m−nm+n1+mnm−nm+n)\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⁡mn−arctan⁡m−nm+n=kπ+arctan⁡(m2+n2m2+n2)=kπ+π4\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⁡mn−arctan⁡m−nm+n∈(−π,π)\arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n}\in \left( -\pi ,\pi \right)we have
kπ+π4∈(−π,π),k∈Zk\pi +\dfrac{\pi }{4}\in \left( -\pi ,\pi \right),k\in \mathbb{Z}
Hence we have k = 0.
Hence arctan⁡mn−arctan⁡m−nm+n=π4\arctan \dfrac{m}{n}-\arctan \dfrac{m-n}{m+n}=\dfrac{\pi }{4}
Hence we have
tan⁡(arctan⁡mn−arctan⁡m−nm+n)=tan⁡π4=1\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⁡(nπ+x)=tan⁡x,n∈Z\tan \left( n\pi +x \right)=\tan x,n\in \mathbb{Z}.
Hence if we take tangents on both sides, we get
tan⁡(arctan⁡mn−arctan⁡m−nm+n)=tan⁡(kπ+π4)=tan⁡π4=1\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 (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right). Whenever not explicitly mentioned, we take arctanx in the principal interval. Hence we have
arctan⁡x∈(−π2,π2),arctan⁡y∈(−π2,π2) ⇒arctan⁡x−arctan⁡y∈(−π,π) \begin{aligned} & \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{aligned}
This is why we chose the value of n, so that nπ+arctan⁡x−y1+xyn\pi +\arctan \dfrac{x-y}{1+xy} is within the interval (−π,π)\left( -\pi ,\pi \right). It can also be proved that unique n satisfies this property.