If `omega ne 1` is a cube root of unity, then `(1)/(pi) sin^(-1) [(omega^(73) + omega^(83))+"tan"(5pi)/(4)]` is equal to ___________

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{1}{\pi} \sin^{-1} \left[ \omega^{73} + \omega^{83} + \tan\left(\frac{5\pi}{4}\right) \right] \]

where \(\omega\) is a cube root of unity, and \(\omega \neq 1\).

### Step 1: Understand the properties of \(\omega\)

The cube roots of unity satisfy the following properties:
1. \(\omega^3 = 1\)
2. \(1 + \omega + \omega^2 = 0\)

From the second property, we can derive that:
\[ \omega + \omega^2 = -1 \]

### Step 2: Simplify \(\omega^{73}\) and \(\omega^{83}\)

To simplify \(\omega^{73}\) and \(\omega^{83}\), we can reduce the exponents modulo 3 since \(\omega^3 = 1\).

- For \(\omega^{73}\):
\[ 73 \mod 3 = 1 \quad \Rightarrow \quad \omega^{73} = \omega^1 = \omega \]

- For \(\omega^{83}\):
\[ 83 \mod 3 = 2 \quad \Rightarrow \quad \omega^{83} = \omega^2 \]

### Step 3: Combine the results

Now we can combine the results:
\[ \omega^{73} + \omega^{83} = \omega + \omega^2 \]

Using the property we derived earlier:
\[ \omega + \omega^2 = -1 \]

### Step 4: Evaluate \(\tan\left(\frac{5\pi}{4}\right)\)

Next, we need to evaluate \(\tan\left(\frac{5\pi}{4}\right)\):
\[ \tan\left(\frac{5\pi}{4}\right) = \tan\left(\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 5: Substitute back into the expression

Now substituting back into our original expression:
\[ \frac{1}{\pi} \sin^{-1} \left[ -1 + 1 \right] = \frac{1}{\pi} \sin^{-1}(0) \]

### Step 6: Evaluate \(\sin^{-1}(0)\)

The value of \(\sin^{-1}(0)\) is:
\[ \sin^{-1}(0) = 0 \]

### Step 7: Final result

Thus, we have:
\[ \frac{1}{\pi} \cdot 0 = 0 \]

Therefore, the final answer is:
\[ \boxed{0} \]