If `sintheta+sinphi=aandcostheta+cosphi=b`, then `tan((theta-phi)/(2))=?`

Options

A. `sqrt((a^(2)+b^(2))/(4-a^(2)-b^(2)))`
B. `sqrt((4-a^(2)-b^(2))/(a^(2)+b^(2)))`
C. `sqrt((a^(2)+b^(2))/(4+a^(2)+b^(2)))`
D. `sqrt((4+a^(2)+b^(2))/(a^(2)+b^(2)))`

Answer

To solve the problem where \( \sin \theta + \sin \phi = a \) and \( \cos \theta + \cos \phi = b \), we need to find \( \tan\left(\frac{\theta - \phi}{2}\right) \).

### Step-by-Step Solution:

1. **Set Up the Equations**:
We have two equations:
\[ \sin \theta + \sin \phi = a \quad \text{(1)} \]
\[ \cos \theta + \cos \phi = b \quad \text{(2)} \]

2. **Square Both Equations**:
Square both equations to eliminate the sine and cosine terms:
\[ (\sin \theta + \sin \phi)^2 = a^2 \implies \sin^2 \theta + \sin^2 \phi + 2 \sin \theta \sin \phi = a^2 \quad \text{(3)} \]
\[ (\cos \theta + \cos \phi)^2 = b^2 \implies \cos^2 \theta + \cos^2 \phi + 2 \cos \theta \cos \phi = b^2 \quad \text{(4)} \]

3. **Add the Two Squared Equations**:
Add equations (3) and (4):
\[ (\sin^2 \theta + \cos^2 \theta) + (\sin^2 \phi + \cos^2 \phi) + 2(\sin \theta \sin \phi + \cos \theta \cos \phi) = a^2 + b^2 \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 1 + 1 + 2(\sin \theta \sin \phi + \cos \theta \cos \phi) = a^2 + b^2 \]
Simplifying gives:
\[ 2 + 2(\sin \theta \sin \phi + \cos \theta \cos \phi) = a^2 + b^2 \]

4. **Use the Cosine Addition Formula**:
The expression \( \sin \theta \sin \phi + \cos \theta \cos \phi \) can be rewritten using the cosine of the difference:
\[ \sin \theta \sin \phi + \cos \theta \cos \phi = \cos(\theta - \phi) \]
Therefore, we have:
\[ 2 + 2\cos(\theta - \phi) = a^2 + b^2 \]
Rearranging gives:
\[ \cos(\theta - \phi) = \frac{a^2 + b^2 - 2}{2} \]

5. **Find \( \tan\left(\frac{\theta - \phi}{2}\right) \)**:
We can use the half-angle formula:
\[ \tan\left(\frac{\theta - \phi}{2}\right) = \sqrt{\frac{1 - \cos(\theta - \phi)}{1 + \cos(\theta - \phi)}} \]
Substituting for \( \cos(\theta - \phi) \):
\[ \tan\left(\frac{\theta - \phi}{2}\right) = \sqrt{\frac{1 - \frac{a^2 + b^2 - 2}{2}}{1 + \frac{a^2 + b^2 - 2}{2}} = \sqrt{\frac{2 - (a^2 + b^2 - 2)}{2 + (a^2 + b^2 - 2)}} \]
Simplifying further:
\[ = \sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}} \]

### Final Answer:
Thus, we have:
\[ \tan\left(\frac{\theta - \phi}{2}\right) = \sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}} \]