If `cos(A+B)=0andsin(A-B)=(sqrt(3))/(2)`then find the value of `tan(A-3B)`.

Answer

To solve the problem step by step, we will use the given equations and trigonometric identities.

### Step 1: Analyze the given equations
We have two equations:
1. \( \cos(A + B) = 0 \)
2. \( \sin(A - B) = \frac{\sqrt{3}}{2} \)

### Step 2: Solve for \( A + B \)
Since \( \cos(A + B) = 0 \), we know that:
\[ A + B = 90^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]
For simplicity, we can take \( n = 0 \):
\[ A + B = 90^\circ \]

### Step 3: Solve for \( A - B \)
Since \( \sin(A - B) = \frac{\sqrt{3}}{2} \), we know that:
\[ A - B = 60^\circ + m \cdot 360^\circ \quad (m \in \mathbb{Z}) \]
For simplicity, we can take \( m = 0 \):
\[ A - B = 60^\circ \]

### Step 4: Set up the system of equations
Now we have a system of equations:
1. \( A + B = 90^\circ \)
2. \( A - B = 60^\circ \)

### Step 5: Solve for \( A \) and \( B \)
To find \( A \) and \( B \), we can add the two equations:
\[ (A + B) + (A - B) = 90^\circ + 60^\circ \]
This simplifies to:
\[ 2A = 150^\circ \implies A = 75^\circ \]

Now, substitute \( A \) back into one of the equations to find \( B \):
\[ 75^\circ + B = 90^\circ \implies B = 90^\circ - 75^\circ = 15^\circ \]

### Step 6: Find \( \tan(A - 3B) \)
Now we need to calculate \( \tan(A - 3B) \):
\[ A - 3B = 75^\circ - 3 \cdot 15^\circ = 75^\circ - 45^\circ = 30^\circ \]
Thus:
\[ \tan(A - 3B) = \tan(30^\circ) \]

### Step 7: Calculate \( \tan(30^\circ) \)
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Final Answer
Therefore, the value of \( \tan(A - 3B) \) is:
\[ \frac{1}{\sqrt{3}} \]

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