On sums involving divisor function, Euler's totient function, and floor function

Every positive integer $l \in \mathbb{N}$ can be formed $l = (m + n)d$, provided $gcd(m,n)=1$. From this point of view, the next formulas $n=\sum_{d|l} \varphi(d)$ and $\frac{n(n+1)}{2}=\sum_{k=1}^{n} \varphi(k)[\frac{n}{k}]$, and these equivalence had been proved. In this paper on an extension of these results, the next identity is proved: $\sum_{k=1}^{n} \sum_{\substack{(a+b)c=k \\ gcd(a,b)=1}} f(a,b)\cdot g(c) = \sum_{k=1}^{n} \sum_{\substack{a+b=k \\ gcd(a,b)=1}} f(a,b) \sum_{i\leq [\frac{n}{k}]} g(i) = \sum_{a+b \leq n} f(\frac{a}{gcd(a,b)},\frac{b}{gcd(a,b)})\cdot g(gcd(a,b))$. We also show the next formulas are corollaries of it: $\sum_{k=1}^{n} \tau(k)=\sum_{k=1}^{n} [\frac{n}{k}] = \sum_{a+b \leq n} \frac{1}{\varphi(\frac{a+b}{gcd(a,b)})}$, $\sum_{d|n} f(d)\cdot g(\frac{n}{d}) = \sum_{k = 1}^{n} f(gcd(k,n))\cdot\frac{g(\frac{n}{gcd(k,n)})}{\varphi(\frac{n}{gcd(k,n)})}$, $\tau(n)=\sum_{a+b = n} \frac{1}{\varphi(\frac{a+b}{gcd(a,b)})}$, $\sum_{\substack{a+b=n \\ gcd(a,b)=1}} gcd(a-1,b+1) = \sum_{a+b=n} \frac{\varphi(n)}{\varphi(\frac{n}{gcd(a,b)})}$, and so on. In addition to it, we evaluate a sequence $\sum_{k=1}^{n} \varphi(k)\tau(k)$.