If in a triangle ABC, `cos A sin B= sin C`, then the value of `tan""A/2 " when " 3b-5c=0` is

Options

A. `1/2`
B. `3/4`
C. `1/sqrt3`
D. `0.33`

Answer

To solve the problem, we need to find the value of \( \tan \frac{A}{2} \) given that \( \cos A \sin B = \sin C \) and \( 3B - 5C = 0 \).

### Step-by-Step Solution:

1. **Use the given condition**:
We know that \( 3B - 5C = 0 \). This implies:
\[ 3B = 5C \quad \text{or} \quad \frac{B}{C} = \frac{5}{3} \]

2. **Apply the sine rule**:
According to the sine rule, we have:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \quad (\text{some constant } k) \]
From this, we can express \( \sin B \) and \( \sin C \) in terms of \( k \):
\[ \sin B = kb \quad \text{and} \quad \sin C = kc \]

3. **Relate \( \sin B \) and \( \sin C \)**:
Since \( \frac{B}{C} = \frac{5}{3} \), we can write:
\[ \frac{\sin B}{\sin C} = \frac{5}{3} \]
Substituting the expressions from the sine rule:
\[ \frac{kb}{kc} = \frac{5}{3} \quad \Rightarrow \quad \frac{b}{c} = \frac{5}{3} \]

4. **Substitute into the equation**:
We are given that \( \cos A \sin B = \sin C \). Substituting for \( \sin B \) and \( \sin C \):
\[ \cos A \cdot kb = kc \]
Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ \cos A \cdot b = c \]

5. **Express \( \cos A \)**:
From the previous step, we can express \( \cos A \):
\[ \cos A = \frac{c}{b} \]

6. **Substitute the ratio**:
We know \( \frac{b}{c} = \frac{5}{3} \), which gives us:
\[ \frac{c}{b} = \frac{3}{5} \]
Thus, we have:
\[ \cos A = \frac{3}{5} \]

7. **Use the half-angle formula**:
We will use the half-angle formula for cosine:
\[ \cos A = \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}} \]
Setting \( \cos A = \frac{3}{5} \):
\[ \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}} = \frac{3}{5} \]

8. **Cross-multiply and simplify**:
Cross-multiplying gives:
\[ 5(1 - \tan^2 \frac{A}{2}) = 3(1 + \tan^2 \frac{A}{2}) \]
Expanding both sides:
\[ 5 - 5\tan^2 \frac{A}{2} = 3 + 3\tan^2 \frac{A}{2} \]
Rearranging terms:
\[ 5 - 3 = 5\tan^2 \frac{A}{2} + 3\tan^2 \frac{A}{2} \]
\[ 2 = 8\tan^2 \frac{A}{2} \]

9. **Solve for \( \tan^2 \frac{A}{2} \)**:
Dividing both sides by 8:
\[ \tan^2 \frac{A}{2} = \frac{1}{4} \]
Taking the square root gives:
\[ \tan \frac{A}{2} = \frac{1}{2} \]

### Final Answer:
Thus, the value of \( \tan \frac{A}{2} \) is \( \frac{1}{2} \).