If `sectheta+tantheta=sqrt3`, then the positive value of `sintheta` is

Options

A. 0
B. `(1)/(2)`
C. `(sqrt3)/(2)`
D. 1

Answer

To solve the equation \( \sec \theta + \tan \theta = \sqrt{3} \) and find the positive value of \( \sin \theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sec \theta + \tan \theta = \sqrt{3} \]

### Step 2: Use the identity
Recall the identity:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
This can be factored as:
\[ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1 \]

### Step 3: Substitute the known value
Since we know \( \sec \theta + \tan \theta = \sqrt{3} \), we can substitute this into the identity:
\[ \sqrt{3}(\sec \theta - \tan \theta) = 1 \]

### Step 4: Solve for \( \sec \theta - \tan \theta \)
Now, we can isolate \( \sec \theta - \tan \theta \):
\[ \sec \theta - \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 5: Set up a system of equations
Now we have two equations:
1. \( \sec \theta + \tan \theta = \sqrt{3} \)
2. \( \sec \theta - \tan \theta = \frac{1}{\sqrt{3}} \)

Let’s denote:
- \( x = \sec \theta \)
- \( y = \tan \theta \)

We can express the two equations as:
1. \( x + y = \sqrt{3} \)
2. \( x - y = \frac{1}{\sqrt{3}} \)

### Step 6: Solve for \( x \) and \( y \)
Add the two equations:
\[ (x + y) + (x - y) = \sqrt{3} + \frac{1}{\sqrt{3}} \]
This simplifies to:
\[ 2x = \sqrt{3} + \frac{1}{\sqrt{3}} \]
Calculating the right side:
\[ \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]
Thus:
\[ 2x = \frac{4}{\sqrt{3}} \implies x = \frac{2}{\sqrt{3}} \]

Now substitute \( x \) back to find \( y \):
\[ \frac{2}{\sqrt{3}} + y = \sqrt{3} \implies y = \sqrt{3} - \frac{2}{\sqrt{3}} = \frac{3 - 2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 7: Find \( \sin \theta \)
We know:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
From \( \sec \theta = \frac{2}{\sqrt{3}} \):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
And from \( \tan \theta = \frac{1}{\sqrt{3}} \):
\[ \sin \theta = \tan \theta \cdot \cos \theta = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} \]

### Final Answer
Thus, the positive value of \( \sin \theta \) is:
\[ \sin \theta = \frac{1}{2} \]