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

Options

A. `pi/3`
B. `pi/6`
C. `-pi/6`
D. `(5pi)/6`

Explanation

To solve the equation \( \sqrt{3} + i = (a + ib)(c + id) \), we will proceed step by step.

### Step 1: Expand the right-hand side
We start by expanding the product on the right-hand side:
\[ (a + ib)(c + id) = ac + i(ad + bc) - bd \]
This can be rewritten as:
\[ (ac - bd) + i(ad + bc) \]

### Step 2: Compare real and imaginary parts
Now, we can compare the real and imaginary parts from both sides of the equation:
\[ \text{Real part: } ac - bd = \sqrt{3} \]
\[ \text{Imaginary part: } ad + bc = 1 \]

### Step 3: Use the tangent addition formula
We need to find the value of \( \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) \). We can use the formula for the sum of two arctangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \]
where \( x = \frac{b}{a} \) and \( y = \frac{d}{c} \).

### Step 4: Substitute the values
Substituting \( x \) and \( y \) into the formula gives us:
\[ \tan^{-1}\left(\frac{\frac{b}{a} + \frac{d}{c}}{1 - \frac{bd}{ac}}\right) \]
This simplifies to:
\[ \tan^{-1}\left(\frac{bc + ad}{ac - bd}\right) \]

### Step 5: Substitute known values
From our earlier comparisons, we know:
- \( ad + bc = 1 \)
- \( ac - bd = \sqrt{3} \)

Thus, we can substitute these values into our expression:
\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Step 6: Evaluate the arctangent
We know that:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Therefore:
\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

### Final Answer
Thus, the value of \( \tan^{-1}\left(\frac{b}{a}\right) + \tan^{-1}\left(\frac{d}{c}\right) \) is:
\[ \frac{\pi}{6} \]