If `cos(theta - theta_(1)) =1/(2sqrt(3))` and `sin(theta-theta_(2))=1/(3sqrt(2))`, where `0 lt theta - theta_(1), theta-theta_(2) lt pi/2`, then the value of `108/5{ cos^(2) (theta_(1)-theta_(2)) + sqrt(6)/18 sin(theta_(1)-theta_(2))}` is …………..

Answer

To solve the problem step by step, we will follow the given information and use trigonometric identities.

### Step 1: Given Information
We have:
1. \( \cos(\theta - \theta_1) = \frac{1}{2\sqrt{3}} \)
2. \( \sin(\theta - \theta_2) = \frac{1}{3\sqrt{2}} \)

### Step 2: Find \( \sin(\theta - \theta_1) \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin(\theta - \theta_1) = \sqrt{1 - \cos^2(\theta - \theta_1)} = \sqrt{1 - \left(\frac{1}{2\sqrt{3}}\right)^2} \]
Calculating \( \cos^2(\theta - \theta_1) \):
\[ \cos^2(\theta - \theta_1) = \frac{1}{12} \]
Thus,
\[ \sin(\theta - \theta_1) = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{2\sqrt{3}} \]

### Step 3: Find \( \cos(\theta - \theta_2) \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \cos(\theta - \theta_2) = \sqrt{1 - \sin^2(\theta - \theta_2)} = \sqrt{1 - \left(\frac{1}{3\sqrt{2}}\right)^2} \]
Calculating \( \sin^2(\theta - \theta_2) \):
\[ \sin^2(\theta - \theta_2) = \frac{1}{18} \]
Thus,
\[ \cos(\theta - \theta_2) = \sqrt{1 - \frac{1}{18}} = \sqrt{\frac{17}{18}} = \frac{\sqrt{17}}{3\sqrt{2}} \]

### Step 4: Find \( \cos(\theta_1 - \theta_2) \)
Using the cosine subtraction formula:
\[ \cos(\theta_1 - \theta_2) = \cos(\theta - \theta_2) \cos(\theta - \theta_1) + \sin(\theta - \theta_2) \sin(\theta - \theta_1) \]
Substituting the values:
\[ \cos(\theta_1 - \theta_2) = \left(\frac{\sqrt{17}}{3\sqrt{2}}\right) \left(\frac{1}{2\sqrt{3}}\right) + \left(\frac{1}{3\sqrt{2}}\right) \left(\frac{\sqrt{11}}{2\sqrt{3}}\right) \]
Calculating:
\[ = \frac{\sqrt{17}}{6\sqrt{6}} + \frac{\sqrt{11}}{6\sqrt{6}} = \frac{\sqrt{17} + \sqrt{11}}{6\sqrt{6}} \]

### Step 5: Find \( \sin(\theta_1 - \theta_2) \)
Using the sine subtraction formula:
\[ \sin(\theta_1 - \theta_2) = \sin(\theta - \theta_2) \cos(\theta - \theta_1) - \cos(\theta - \theta_2) \sin(\theta - \theta_1) \]
Substituting the values:
\[ \sin(\theta_1 - \theta_2) = \left(\frac{1}{3\sqrt{2}}\right) \left(\frac{1}{2\sqrt{3}}\right) - \left(\frac{\sqrt{17}}{3\sqrt{2}}\right) \left(\frac{\sqrt{11}}{2\sqrt{3}}\right) \]
Calculating:
\[ = \frac{1}{6\sqrt{6}} - \frac{\sqrt{187}}{6\sqrt{6}} = \frac{1 - \sqrt{187}}{6\sqrt{6}} \]

### Step 6: Calculate the Final Expression
Now we substitute \( \cos^2(\theta_1 - \theta_2) \) and \( \sin(\theta_1 - \theta_2) \) into the expression:
\[ \frac{108}{5} \left( \left(\frac{\sqrt{17} + \sqrt{11}}{6\sqrt{6}}\right)^2 + \frac{\sqrt{6}}{18} \left(\frac{1 - \sqrt{187}}{6\sqrt{6}}\right) \right) \]
Calculating \( \cos^2(\theta_1 - \theta_2) \):
\[ = \frac{(17 + 11 + 2\sqrt{187})}{216} \]
And simplifying the entire expression gives us the final value.

### Final Answer
After performing all calculations, we find that the value of the expression is \( 3 \).