Domain of $f(x)=\sin^{-1}\left(\frac{[x]}{\{x\}}\right)$ where [.] and {.} denote the greatest integer function and fractional part respectively, is

(0, 1)

We must have, $-1≤\frac{[x]}{\{x\}}≤1$

From $-1≤\frac{[x]}{\{x\}}$, we get $\frac{\{x\}-[x]}{\{x\}}≥0⇒\frac{x}{\{x\}}>0⇒x∈(0,∞)∼I^+$

From $\frac{[x]}{\{x\}}≤1$ we get, $\frac{[x]-\{x\}}{\{x\}}≤0⇒[x]≤\{x\}$ where $\{x\} ≠ 0$

$⇒x∈(0,∞)∪(0,1)$. Thus domain is (0, 1)