If ` cos x-(cot beta sinx )/( 2 )=sqrt(3)/(2)` , then the value of ` tan""(x)/(2)` is

A. `tan""(beta)/(2) tan 15^(@)`
B. `tan""(beta)/(2)`
C.`tan 15^(@)`
D. None of these

Options

A. `tan""(beta)/(2) tan 15^(@)`
B. `tan""(beta)/(2)`
C. `tan 15^(@)`
D. None of the above

Answer

To solve the equation \( \cos x - \frac{\cot \beta \sin x}{2} = \frac{\sqrt{3}}{2} \), we will follow these steps:

### Step 1: Use Trigonometric Identities
We will use the half-angle formulas for cosine and sine:
- \( \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)
- \( \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)

Let \( t = \tan \frac{x}{2} \). Then we can rewrite the equation as:
\[ \frac{1 - t^2}{1 + t^2} - \frac{\cot \beta \cdot \frac{2t}{1 + t^2}}{2} = \frac{\sqrt{3}}{2} \]

### Step 2: Simplify the Equation
Substituting the sine and cosine values into the equation gives:
\[ \frac{1 - t^2}{1 + t^2} - \frac{\cot \beta \cdot t}{1 + t^2} = \frac{\sqrt{3}}{2} \]
Now, we can multiply through by \( 1 + t^2 \) to eliminate the denominator:
\[ 1 - t^2 - \frac{\cot \beta \cdot t}{2} = \frac{\sqrt{3}}{2}(1 + t^2) \]

### Step 3: Rearranging the Equation
Rearranging this gives:
\[ 1 - t^2 - \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} t^2 - \frac{\cot \beta \cdot t}{2} = 0 \]
Combining like terms results in:
\[ (1 - \frac{\sqrt{3}}{2}) - (1 + \frac{\sqrt{3}}{2})t^2 - \frac{\cot \beta \cdot t}{2} = 0 \]

### Step 4: Form a Quadratic Equation
Let’s denote \( p = t \) (where \( t = \tan \frac{x}{2} \)):
\[ -\left(1 + \frac{\sqrt{3}}{2}\right)p^2 - \frac{\cot \beta}{2} p + \left(1 - \frac{\sqrt{3}}{2}\right) = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = -(1 + \frac{\sqrt{3}}{2}) \)
- \( b = -\frac{\cot \beta}{2} \)
- \( c = 1 - \frac{\sqrt{3}}{2} \)

Substituting these values into the quadratic formula will yield the values of \( p \).

### Step 6: Find \( \tan \frac{x}{2} \)
After solving the quadratic equation, we can express \( \tan \frac{x}{2} \) in terms of \( \tan \frac{\beta}{2} \) and \( \tan 15^\circ \).

### Conclusion
After simplification, we find:
\[ \tan \frac{x}{2} = \tan \frac{\beta}{2} \tan 15^\circ \]
Thus, the answer is:
**Option A: \( \tan \frac{\beta}{2} \tan 15^\circ \)**

---