If `sqrt(3)+i=(a+i b)(c+i d)` , then find the value of `tan^(-1)(b//a)tan^(-1)(d//c)dot`

Explanation

To solve the problem, we start with the equation given:

\[ \sqrt{3} + i = (a + ib)(c + id) \]

### Step 1: Expand the Right Side
We expand the right side using the distributive property:

\[ (a + ib)(c + id) = ac + iad + ibc + i^2bd \]

Since \(i^2 = -1\), we can rewrite this as:

\[ ac + i(ad + bc - bd) \]

### Step 2: Separate Real and Imaginary Parts
Now we can separate the real and imaginary parts:

\[ \text{Real part: } ac - bd \]
\[ \text{Imaginary part: } ad + bc \]

### Step 3: Set Up Equations
Now we can set up equations by comparing the real and imaginary parts with \(\sqrt{3} + i\):

1. \(ac - bd = \sqrt{3}\)
2. \(ad + bc = 1\)

### Step 4: Solve for \(b/a\) and \(d/c\)
We need to find \( \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) \). We can use the formula for the sum of arctangents:

\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]

Let \(x = \frac{b}{a}\) and \(y = \frac{d}{c}\). Therefore, we have:

\[ \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) = \tan^{-1}\left(\frac{\frac{b}{a} + \frac{d}{c}}{1 - \frac{b}{a} \cdot \frac{d}{c}}\right) \]

### Step 5: Substitute Values
Substituting \(x\) and \(y\) into the formula gives:

\[ \tan^{-1}\left(\frac{\frac{bc + ad}{ac}}{1 - \frac{bd}{ac}}\right) \]

### Step 6: Simplify
Now we substitute the values from our earlier equations:

- \(bc + ad = 1\)
- \(ac - bd = \sqrt{3}\)

Thus, we can write:

\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Step 7: Find the Angle
We know that:

\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]

Thus:

\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

### Final Answer
Therefore, the final answer is:

\[ \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

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