If `alpha` and `beta` are roots of equation `3/4"sin"((theta)/9)=sin^(3)theta+3sin^(3)((theta)/3)+9sin^(3)((theta)/9)+1/(4sqrt(2))` for `0lt theta lt (pi)/2`, then `tanalpha+tanbeta` is equal to

Options

A. `2+sqrt(3)`
B. `3+sqrt(3)`
C. `3-sqrt(3)`
D. `2-sqrt(3)`

Explanation

To solve the equation given in the problem, we will follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ \frac{3}{4} \sin\left(\frac{\theta}{9}\right) = \sin^3(\theta) + 3\sin^3\left(\frac{\theta}{3}\right) + 9\sin^3\left(\frac{\theta}{9}\right) + \frac{1}{4\sqrt{2}} \]

### Step 2: Use the identity for \(\sin(3\theta)\)
We know that:
\[ \frac{1}{\sin(3\theta)} = \frac{3}{4} \sin\left(\frac{\theta}{9}\right) - \sin^3(\theta) - 3\sin^3\left(\frac{\theta}{3}\right) - 9\sin^3\left(\frac{\theta}{9}\right) \]
This allows us to rewrite our equation in terms of \(\sin(3\theta)\).

### Step 3: Rearranging the equation
Rearranging gives us:
\[ \frac{1}{4\sin(3\theta)} = \frac{1}{4\sqrt{2}} \]
Multiplying both sides by 4 results in:
\[ \frac{1}{\sin(3\theta)} = \sqrt{2} \]

### Step 4: Solve for \(3\theta\)
Taking the reciprocal gives:
\[ \sin(3\theta) = \frac{1}{\sqrt{2}} \]
The angles for which \(\sin\) equals \(\frac{1}{\sqrt{2}}\) are:
\[ 3\theta = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad 3\theta = \frac{3\pi}{4} + 2k\pi \]
for integers \(k\).

### Step 5: Solve for \(\theta\)
Dividing by 3 gives:
\[ \theta = \frac{\pi}{12} + \frac{2k\pi}{3} \quad \text{or} \quad \theta = \frac{\pi}{4} + \frac{2k\pi}{3} \]
Considering the range \(0 < \theta < \frac{\pi}{2}\), we find:
- For \(k = 0\): \(\theta = \frac{\pi}{12}\) and \(\theta = \frac{\pi}{4}\).

### Step 6: Identify roots \(\alpha\) and \(\beta\)
Thus, the roots are:
\[ \alpha = \frac{\pi}{12}, \quad \beta = \frac{\pi}{4} \]

### Step 7: Calculate \(\tan\alpha + \tan\beta\)
Now we compute:
\[ \tan\alpha = \tan\left(\frac{\pi}{12}\right) \quad \text{and} \quad \tan\beta = \tan\left(\frac{\pi}{4}\right) \]
We know:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Using the tangent half-angle formula:
\[ \tan\left(\frac{\pi}{12}\right) = 2 - \sqrt{3} \]
Thus:
\[ \tan\alpha + \tan\beta = (2 - \sqrt{3}) + 1 = 3 - \sqrt{3} \]

### Final Answer
Therefore, the value of \(\tan\alpha + \tan\beta\) is:
\[ \boxed{3 - \sqrt{3}} \]