If `0ltAltBltpi, sin A-sinB=(1)/(sqrt2) and cos A-cos B=sqrt((3)/(2))`, then the value of `A+B` is equal to

Options

A. `(2pi)/(3)`
B. `(5pi)/(6)`
C. `pi`
D. `(4pi)/(3)`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
\[ \sin A - \sin B = \frac{1}{\sqrt{2}} \quad \text{(Equation 1)} \]
\[ \cos A - \cos B = \sqrt{\frac{3}{2}} \quad \text{(Equation 2)} \]

2. **Using the Sine Difference Formula:**
We can use the sine difference formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Setting this equal to \(\frac{1}{\sqrt{2}}\):
\[ 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \]

3. **Using the Cosine Difference Formula:**
Similarly, for cosine:
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Setting this equal to \(\sqrt{\frac{3}{2}}\):
\[ -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) = \sqrt{\frac{3}{2}} \]

4. **Dividing the Two Equations:**
Now, we divide Equation 2 by Equation 1:
\[ \frac{-2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)}{2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)} = \frac{\sqrt{\frac{3}{2}}}{\frac{1}{\sqrt{2}}} \]
This simplifies to:
\[ -\tan\left(\frac{A+B}{2}\right) = \sqrt{3} \]
Thus:
\[ \tan\left(\frac{A+B}{2}\right) = -\sqrt{3} \]

5. **Finding the Angle:**
The value of \(\tan\theta = -\sqrt{3}\) corresponds to angles in the second quadrant. The reference angle for \(\tan\theta = \sqrt{3}\) is \(\frac{\pi}{3}\). Therefore:
\[ \frac{A+B}{2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

6. **Calculating \(A + B\):**
Multiplying both sides by 2 gives:
\[ A + B = 2 \times \frac{2\pi}{3} = \frac{4\pi}{3} \]

Thus, the final answer is:
\[ \boxed{\frac{4\pi}{3}} \]