If `0ltAltBltpi,sinA-sinB=(1)/(sqrt(2)),cosA-cosB=sqrt((3)/(2))`, then `A+B=`

Options

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

Answer

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

1. \( \sin A - \sin B = \frac{1}{\sqrt{2}} \)
2. \( \cos A - \cos B = \sqrt{\frac{3}{2}} \)

We will square both equations and add them together.

### Step 1: Square both equations

Squaring the first equation:
\[ (\sin A - \sin B)^2 = \left(\frac{1}{\sqrt{2}}\right)^2 \]
This simplifies to:
\[ \sin^2 A - 2 \sin A \sin B + \sin^2 B = \frac{1}{2} \]

Squaring the second equation:
\[ (\cos A - \cos B)^2 = \left(\sqrt{\frac{3}{2}}\right)^2 \]
This simplifies to:
\[ \cos^2 A - 2 \cos A \cos B + \cos^2 B = \frac{3}{2} \]

### Step 2: Add the two squared equations

Now, we add the two equations:
\[ \sin^2 A + \cos^2 A - 2 \sin A \sin B + \sin^2 B + \cos^2 B - 2 \cos A \cos B = \frac{1}{2} + \frac{3}{2} \]
Using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \) and \( \sin^2 B + \cos^2 B = 1 \):
\[ 1 + 1 - 2(\sin A \sin B + \cos A \cos B) = 2 \]
This simplifies to:
\[ 2 - 2(\sin A \sin B + \cos A \cos B) = 2 \]

### Step 3: Simplify the equation

Rearranging gives:
\[ -2(\sin A \sin B + \cos A \cos B) = 0 \]
Thus:
\[ \sin A \sin B + \cos A \cos B = 0 \]
This can be rewritten using the cosine of the difference formula:
\[ \cos(A - B) = 0 \]

### Step 4: Solve for \( A - B \)

The equation \( \cos(A - B) = 0 \) implies:
\[ A - B = \frac{\pi}{2} \quad \text{or} \quad A - B = \frac{3\pi}{2} \]
Since \( 0 < A < B < \pi \), we take:
\[ A - B = \frac{\pi}{2} \quad \Rightarrow \quad B = A + \frac{\pi}{2} \]

### Step 5: Substitute \( B \) into the sine equation

Substituting \( B = A + \frac{\pi}{2} \) into the first equation:
\[ \sin A - \sin\left(A + \frac{\pi}{2}\right) = \frac{1}{\sqrt{2}} \]
Using the identity \( \sin\left(A + \frac{\pi}{2}\right) = \cos A \):
\[ \sin A - \cos A = \frac{1}{\sqrt{2}} \]

### Step 6: Substitute \( B \) into the cosine equation

Substituting \( B = A + \frac{\pi}{2} \) into the second equation:
\[ \cos A - \cos\left(A + \frac{\pi}{2}\right) = \sqrt{\frac{3}{2}} \]
Using the identity \( \cos\left(A + \frac{\pi}{2}\right) = -\sin A \):
\[ \cos A + \sin A = \sqrt{\frac{3}{2}} \]

### Step 7: Solve the system of equations

Now we have the system:
1. \( \sin A - \cos A = \frac{1}{\sqrt{2}} \)
2. \( \sin A + \cos A = \sqrt{\frac{3}{2}} \)

Adding these two equations:
\[ 2 \sin A = \frac{1}{\sqrt{2}} + \sqrt{\frac{3}{2}} \]
Thus:
\[ \sin A = \frac{\frac{1}{\sqrt{2}} + \sqrt{\frac{3}{2}}}{2} \]

### Step 8: Find \( A \) and \( B \)

Calculating \( A \):
Using known values, we find:
\[ A = 75^\circ = \frac{5\pi}{12} \]
Then:
\[ B = A + \frac{\pi}{2} = \frac{5\pi}{12} + \frac{6\pi}{12} = \frac{11\pi}{12} \]

### Step 9: Find \( A + B \)

Finally, we calculate:
\[ A + B = \frac{5\pi}{12} + \frac{11\pi}{12} = \frac{16\pi}{12} = \frac{4\pi}{3} \]

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