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 will use the given equations and trigonometric identities.

### Step 1: Use the formulas for sine and cosine differences
We know that:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]

### Step 2: Set up the equations
From the problem, we have:
\[ \sin A - \sin B = \frac{1}{\sqrt{2}} \]
\[ \cos A - \cos B = \sqrt{\frac{3}{2}} \]

Substituting the formulas into these equations gives us:
\[ 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{2}} \tag{1} \]
\[ -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) = \sqrt{\frac{3}{2}} \tag{2} \]

### Step 3: Divide equation (1) by equation (2)
Dividing equation (1) by equation (2):
\[ \frac{2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)}{-2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{\frac{3}{2}}} \]
This simplifies to:
\[ -\cot\left(\frac{A+B}{2}\right) = \frac{1/\sqrt{2}}{\sqrt{3}/\sqrt{2}} = \frac{1}{\sqrt{3}} \]

### Step 4: Solve for \(\frac{A+B}{2}\)
Thus, we have:
\[ \cot\left(\frac{A+B}{2}\right) = -\frac{1}{\sqrt{3}} \]
The value of \(\cot\) is \(-\frac{1}{\sqrt{3}}\) at angles in the second quadrant. The reference angle for \(\cot^{-1}\left(\frac{1}{\sqrt{3}}\right)\) is \(\frac{\pi}{3}\), so:
\[ \frac{A+B}{2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

### Step 5: Find \(A+B\)
Multiplying both sides by 2 gives:
\[ A + B = 2 \times \frac{2\pi}{3} = \frac{4\pi}{3} \]

### Final Answer
Thus, the value of \(A + B\) is:
\[ \boxed{\frac{4\pi}{3}} \]