If `P+Q=(7pi)/(6)`, then find the value of `(sqrt(3)+tanP)xx(sqrt(3)+tanQ)`.

Explanation

To solve the problem, we need to find the value of \((\sqrt{3} + \tan P)(\sqrt{3} + \tan Q)\) given that \(P + Q = \frac{7\pi}{6}\).

### Step 1: Write down the given equation
We start with the equation:
\[ P + Q = \frac{7\pi}{6} \]

### Step 2: Apply the tangent function
Taking the tangent of both sides, we have:
\[ \tan(P + Q) = \tan\left(\frac{7\pi}{6}\right) \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan(P + Q) = \frac{\tan P + \tan Q}{1 - \tan P \tan Q} \]
So we can write:
\[ \frac{\tan P + \tan Q}{1 - \tan P \tan Q} = \tan\left(\frac{7\pi}{6}\right) \]

### Step 4: Calculate \(\tan\left(\frac{7\pi}{6}\right)\)
We know that:
\[ \tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus, we have:
\[ \frac{\tan P + \tan Q}{1 - \tan P \tan Q} = \frac{1}{\sqrt{3}} \]

### Step 5: Cross-multiply
Cross-multiplying gives us:
\[ \sqrt{3}(\tan P + \tan Q) = 1 - \tan P \tan Q \]
This can be rearranged to form our first equation:
\[ \sqrt{3}\tan P + \sqrt{3}\tan Q + \tan P \tan Q = 1 \quad \text{(Equation 1)} \]

### Step 6: Expand the expression
Now we need to evaluate:
\[ (\sqrt{3} + \tan P)(\sqrt{3} + \tan Q) \]
Expanding this gives:
\[ = \sqrt{3} \cdot \sqrt{3} + \sqrt{3} \tan Q + \tan P \sqrt{3} + \tan P \tan Q \]
\[ = 3 + \sqrt{3}(\tan P + \tan Q) + \tan P \tan Q \]

### Step 7: Substitute from Equation 1
From Equation 1, we know that:
\[ \sqrt{3}(\tan P + \tan Q) + \tan P \tan Q = 1 \]
Substituting this into our expression gives:
\[ = 3 + 1 = 4 \]

### Final Answer
Thus, the value of \((\sqrt{3} + \tan P)(\sqrt{3} + \tan Q)\) is:
\[ \boxed{4} \]