If `(rcostheta-sqrt(3))^(2)+(rsintheta-1)^(2)=0` then the value of `(rtantheta+sectheta)/(rsec theta+tantheta)` is equal to

Options

A. `(4)/(5)`
B. `(5)/(4)`
C. `(sqrt(3))/(4)`
D. `(sqrt(5))/(4)`

Answer

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

\[ (r \cos \theta - \sqrt{3})^2 + (r \sin \theta - 1)^2 = 0 \]

### Step 1: Analyze the equation
Since the sum of two squares is equal to zero, each square must be individually equal to zero. Thus, we can set up two equations:

1. \( r \cos \theta - \sqrt{3} = 0 \)
2. \( r \sin \theta - 1 = 0 \)

### Step 2: Solve for \( r \cos \theta \) and \( r \sin \theta \)
From the first equation:

\[ r \cos \theta = \sqrt{3} \implies \cos \theta = \frac{\sqrt{3}}{r} \]

From the second equation:

\[ r \sin \theta = 1 \implies \sin \theta = \frac{1}{r} \]

### Step 3: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \left(\frac{1}{r}\right)^2 + \left(\frac{\sqrt{3}}{r}\right)^2 = 1 \]

This simplifies to:

\[ \frac{1}{r^2} + \frac{3}{r^2} = 1 \]

\[ \frac{4}{r^2} = 1 \implies r^2 = 4 \implies r = 2 \]

### Step 4: Find \( \theta \)
Now substituting \( r = 2 \) back into the equations for \( \sin \theta \) and \( \cos \theta \):

\[ \sin \theta = \frac{1}{2} \quad \text{and} \quad \cos \theta = \frac{\sqrt{3}}{2} \]

This corresponds to \( \theta = 30^\circ \) (or \( \frac{\pi}{6} \) radians).

### Step 5: Calculate \( \frac{r \tan \theta + \sec \theta}{r \sec \theta + \tan \theta} \)
Now we can substitute \( r = 2 \) and \( \theta = 30^\circ \):

- \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)
- \( \sec 30^\circ = \frac{2}{\sqrt{3}} \)

Substituting these values into the expression:

\[ \frac{2 \cdot \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}}{2 \cdot \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}}} \]

### Step 6: Simplify the expression
The numerator becomes:

\[ \frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

The denominator becomes:

\[ \frac{4}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}} \]

Thus, we have:

\[ \frac{\frac{4}{\sqrt{3}}}{\frac{5}{\sqrt{3}}} = \frac{4}{5} \]

### Final Answer
The value of \( \frac{r \tan \theta + \sec \theta}{r \sec \theta + \tan \theta} \) is:

\[ \boxed{\frac{4}{5}} \]