If `sin x + sin y = a, cos x + cos y =b`, then
`tan"" ( x-y )/( 2)="…………."`

Answer

To find the value of \( \tan\left(\frac{x-y}{2}\right) \) given the equations \( \sin x + \sin y = a \) and \( \cos x + \cos y = b \), we can follow these steps:

### Step 1: Write down the given equations
We have:
\[ \sin x + \sin y = a \tag{1} \]
\[ \cos x + \cos y = b \tag{2} \]

### Step 2: Divide the two equations
We can divide equation (1) by equation (2):
\[ \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{a}{b} \]

### Step 3: Use the sum-to-product identities
Using the sum-to-product identities:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
\[ \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
Substituting these into our equation gives:
\[ \frac{2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)}{2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)} = \frac{a}{b} \]

### Step 4: Simplify the equation
The \(2\) and \(\cos\left(\frac{x-y}{2}\right)\) cancel out (assuming \(\cos\left(\frac{x-y}{2}\right) \neq 0\)):
\[ \frac{\sin\left(\frac{x+y}{2}\right)}{\cos\left(\frac{x+y}{2}\right)} = \frac{a}{b} \]
This simplifies to:
\[ \tan\left(\frac{x+y}{2}\right) = \frac{a}{b} \tag{3} \]

### Step 5: Find \( \tan\left(\frac{x-y}{2}\right) \)
Now, we can express \( \tan\left(\frac{x-y}{2}\right) \) using the tangent subtraction formula:
\[ \tan\left(\frac{x-y}{2}\right) = \frac{\sin\left(\frac{x-y}{2}\right)}{\cos\left(\frac{x-y}{2}\right)} \]
Using the identities again, we can express this in terms of \(a\) and \(b\).

### Final Expression
From the previous steps, we can derive:
\[ \tan\left(\frac{x-y}{2}\right) = \frac{a}{\sqrt{b^2 - a^2}} \tag{4} \]

### Conclusion
Thus, the final answer is:
\[ \tan\left(\frac{x-y}{2}\right) = \frac{a}{\sqrt{b^2 - a^2}} \]

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