Find the range of each of the following functions: (where {.} and [.] represents fractional part and greatest integer part functions respectively)
`f(x)=(sinx)/(sqrt(1+tan^(2)x))+(cosx)/(sqrt(1+cot^(2)x))`

Explanation

To find the range of the function \( f(x) = \frac{\sin x}{\sqrt{1 + \tan^2 x}} + \frac{\cos x}{\sqrt{1 + \cot^2 x}} \), we can simplify it step by step.

### Step 1: Simplify the function
We start with the given function:
\[ f(x) = \frac{\sin x}{\sqrt{1 + \tan^2 x}} + \frac{\cos x}{\sqrt{1 + \cot^2 x}} \]

Using the Pythagorean identity, we know:
\[ 1 + \tan^2 x = \sec^2 x \quad \text{and} \quad 1 + \cot^2 x = \csc^2 x \]

Thus, we can rewrite the function:
\[ f(x) = \frac{\sin x}{\sqrt{\sec^2 x}} + \frac{\cos x}{\sqrt{\csc^2 x}} \]

### Step 2: Further simplify
The square roots of the secant and cosecant functions can be simplified:
\[ \sqrt{\sec^2 x} = \sec x = \frac{1}{\cos x} \quad \text{and} \quad \sqrt{\csc^2 x} = \csc x = \frac{1}{\sin x} \]

Substituting these into the function gives:
\[ f(x) = \sin x \cdot \cos x + \cos x \cdot \sin x \]
\[ = \sin x \cdot \cos x + \cos x \cdot \sin x = 2 \sin x \cos x \]

### Step 3: Use the double angle identity
We can use the double angle identity:
\[ 2 \sin x \cos x = \sin(2x) \]

Thus, we have:
\[ f(x) = \sin(2x) \]

### Step 4: Determine the range of \( \sin(2x) \)
The sine function oscillates between -1 and 1 for all real values of its argument. Therefore, the range of \( f(x) = \sin(2x) \) is:
\[ [-1, 1] \]

### Conclusion
The range of the function \( f(x) \) is:
\[ \boxed{[-1, 1]} \]