If `tan(A+B)=sqrt(3) and sqrt(3)tan(A-B)=1`, find the angles A and B,where A and B are Acute Angles.

Answer

To solve the problem, we need to find the angles A and B given the equations:

1. \( \tan(A + B) = \sqrt{3} \)
2. \( \sqrt{3} \tan(A - B) = 1 \)

### Step 1: Analyze the first equation

From the first equation, we have:

\[ \tan(A + B) = \sqrt{3} \]

We know that \( \tan(60^\circ) = \sqrt{3} \). Therefore, we can set:

\[ A + B = 60^\circ \quad \text{(Equation 1)} \]

### Step 2: Analyze the second equation

From the second equation, we can rewrite it as:

\[ \sqrt{3} \tan(A - B) = 1 \]

Dividing both sides by \( \sqrt{3} \):

\[ \tan(A - B) = \frac{1}{\sqrt{3}} \]

We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). Therefore, we can set:

\[ A - B = 30^\circ \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously

Now we have two equations:

1. \( A + B = 60^\circ \)
2. \( A - B = 30^\circ \)

We can add these two equations:

\[ (A + B) + (A - B) = 60^\circ + 30^\circ \]

This simplifies to:

\[ 2A = 90^\circ \]

Dividing both sides by 2 gives:

\[ A = 45^\circ \]

### Step 4: Substitute to find B

Now, we can substitute the value of \( A \) back into Equation 1 to find \( B \):

\[ 45^\circ + B = 60^\circ \]

Subtracting \( 45^\circ \) from both sides gives:

\[ B = 60^\circ - 45^\circ = 15^\circ \]

### Conclusion

Thus, the angles are:

\[ A = 45^\circ \quad \text{and} \quad B = 15^\circ \]

Both angles A and B are acute angles.