If $m$ and $n$ be positive integers Prove that $\varphi(mn)= \varphi(m,n) \cdot \varphi[m,n]$ where [m, n] =l.c.m of $a$ and $b$ And (m, n)=g.c.d of $a$ and $b$
My approach $\varphi (mn)=\varphi ((m, n) [m, n])$ $\implies ({mn/[m, n]} ,[m, n])$ $=({mn/[m, n]},{mn/(m, n)}$ Am I in right direction
As mentioned in the comments, it is not true that $$ \varphi(mn)= \varphi(m,n) \varphi[m,n] $$ For instance, it fails for $m=n=p$, where $p$ is prime, because $\varphi(p^2)=p(p-1) \ne p^2 = \varphi(p)^2$.
The correct identity is $$ \varphi(mn) = \varphi(m) \varphi(n) \frac{d}{\varphi(d)} $$ where $d = \gcd(m,n)$. See a proof here.
This symmetric version of that identity looks much nicer $$ \frac{\varphi(mn)}{mn} \frac{\varphi(d)}{d} = \frac{\varphi(m)}{m} \frac{\varphi(n)}{n} $$