If tan A = p and cot B =q then express sin ( A + B) in terms of p and q.

Answer

To express \( \sin(A + B) \) in terms of \( p \) and \( q \) given that \( \tan A = p \) and \( \cot B = q \), we can follow these steps:

### Step 1: Express \( \tan A \) and \( \cot B \) in terms of sine and cosine
Given:
\[ \tan A = p \implies \frac{\sin A}{\cos A} = p \]
This implies:
\[ \sin A = p \cos A \]

Also, since \( \cot B = q \):
\[ \cot B = \frac{1}{\tan B} = q \implies \tan B = \frac{1}{q} \implies \frac{\sin B}{\cos B} = \frac{1}{q} \]
This implies:
\[ \sin B = \frac{1}{q} \cos B \]

### Step 2: Use the sine addition formula
The sine addition formula states:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

### Step 3: Substitute \( \sin A \) and \( \sin B \) into the formula
Substituting the expressions we found for \( \sin A \) and \( \sin B \):
\[ \sin(A + B) = (p \cos A) \cos B + \cos A \left(\frac{1}{q} \cos B\right) \]

### Step 4: Factor out \( \cos A \cos B \)
\[ \sin(A + B) = \cos A \cos B \left( p + \frac{1}{q} \right) \]

### Step 5: Find \( \cos A \) and \( \cos B \) in terms of \( p \) and \( q \)
From \( \tan A = p \):
\[ \cos^2 A = \frac{1}{1 + p^2} \implies \cos A = \frac{1}{\sqrt{1 + p^2}} \]

From \( \tan B = \frac{1}{q} \):
\[ \cos^2 B = \frac{1}{1 + \left(\frac{1}{q}\right)^2} = \frac{q^2}{1 + q^2} \implies \cos B = \frac{q}{\sqrt{1 + q^2}} \]

### Step 6: Substitute \( \cos A \) and \( \cos B \) back into the sine addition formula
Now substituting \( \cos A \) and \( \cos B \):
\[ \sin(A + B) = \left(\frac{1}{\sqrt{1 + p^2}}\right) \left(\frac{q}{\sqrt{1 + q^2}}\right) \left( p + \frac{1}{q} \right) \]

### Step 7: Simplify the expression
\[ \sin(A + B) = \frac{q(p + \frac{1}{q})}{\sqrt{(1 + p^2)(1 + q^2)}} \]
\[ = \frac{pq + 1}{\sqrt{(1 + p^2)(1 + q^2)}} \]

Thus, the final expression for \( \sin(A + B) \) in terms of \( p \) and \( q \) is:
\[ \sin(A + B) = \frac{pq + 1}{\sqrt{(1 + p^2)(1 + q^2)}} \]