Let a,b, c are the sides opposite to angles A, B , C respectively in a `Delta ABC tan""(A-B)/(2)=(a-b)/(a+b)cot ""C/2 and (a)/(sin A)=(b)/(sin B) =(c)/(sin C),`
If `a=6,b=3 and cos (A-B) =4/5`
Valus of sin A is equal to

Options

A. `(1)/(sqrt5)`
B. `(2)/(sqrt5)`
C. `(1)/(2sqrt5)`
D. `(1)/(sqrt3)`

Answer

To solve the problem, we need to find the value of sin A given the sides of the triangle and the cosine of the angle difference. Let's break it down step by step.

### Step 1: Understand the Given Information
We are given:
- \( a = 6 \)
- \( b = 3 \)
- \( \cos(A - B) = \frac{4}{5} \)

### Step 2: Use the Relationship Involving Tangent
We have the equation:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2} \]
Substituting the values of \( a \) and \( b \):
\[ \tan\left(\frac{A - B}{2}\right) = \frac{6 - 3}{6 + 3} \cot\left(\frac{C}{2}\right) \]
This simplifies to:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{3}{9} \cot\left(\frac{C}{2}\right) = \frac{1}{3} \cot\left(\frac{C}{2}\right) \]

### Step 3: Express \(\tan\left(\frac{A - B}{2}\right)\) in Terms of Cosine
Using the identity:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{\sin(A - B)}{1 + \cos(A - B)} \]
We know:
\[ \cos(A - B) = \frac{4}{5} \]
Thus, we can find \(\sin(A - B)\) using the Pythagorean identity:
\[ \sin^2(A - B) + \cos^2(A - B) = 1 \]
\[ \sin^2(A - B) + \left(\frac{4}{5}\right)^2 = 1 \]
\[ \sin^2(A - B) + \frac{16}{25} = 1 \]
\[ \sin^2(A - B) = 1 - \frac{16}{25} = \frac{9}{25} \]
\[ \sin(A - B) = \frac{3}{5} \]

Now substituting back into the tangent expression:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{\frac{3}{5}}{1 + \frac{4}{5}} = \frac{\frac{3}{5}}{\frac{9}{5}} = \frac{3}{9} = \frac{1}{3} \]

### Step 4: Equate and Solve for \(\cot\left(\frac{C}{2}\right)\)
From the previous step:
\[ \frac{1}{3} \cot\left(\frac{C}{2}\right) = \frac{1}{3} \]
Thus, we have:
\[ \cot\left(\frac{C}{2}\right) = 1 \]
This implies:
\[ \tan\left(\frac{C}{2}\right) = 1 \quad \Rightarrow \quad \frac{C}{2} = \frac{\pi}{4} \quad \Rightarrow \quad C = \frac{\pi}{2} \]

### Step 5: Use the Sine Rule
Since \( C = 90^\circ \), we can use the sine rule:
\[ \frac{a}{\sin A} = \frac{c}{\sin C} \]
Here, \( \sin C = 1 \) (since \( C = 90^\circ \)):
\[ \frac{6}{\sin A} = c \]
We need to find \( c \) using the Pythagorean theorem:
\[ c = \sqrt{a^2 + b^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \]
Thus:
\[ \frac{6}{\sin A} = 3\sqrt{5} \]
Rearranging gives:
\[ \sin A = \frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}} \]

### Final Answer
The value of \( \sin A \) is:
\[ \sin A = \frac{2}{\sqrt{5}} \]