If `p sec theta - b tan theta =a` and `q sec theta +a tan theta =b`, then

Options

A. `a^(2) + b^(2) = p^(2) + q^(2)`
B. `a^(2) + p^(2) = b^(2) + q^(2)`
C. `a^(2) + q^(2) = b^(2) +p^(2)`
D. none

Answer

To solve the given equations \( p \sec \theta - b \tan \theta = a \) and \( q \sec \theta + a \tan \theta = b \), we will derive a relationship between \( A, B, P, \) and \( Q \).

### Step-by-step Solution:

1. **Start with the first equation:**
\[ p \sec \theta - b \tan \theta = a \]
We can express \( \sec \theta \) and \( \tan \theta \) in terms of sine and cosine:
\[ p \frac{1}{\cos \theta} - b \frac{\sin \theta}{\cos \theta} = a \]
This can be rewritten as:
\[ \frac{p - b \sin \theta}{\cos \theta} = a \]

2. **Multiply both sides by \( \cos \theta \):**
\[ p - b \sin \theta = a \cos \theta \]
Rearranging gives us:
\[ p = a \cos \theta + b \sin \theta \quad \text{(Equation 1)} \]

3. **Now consider the second equation:**
\[ q \sec \theta + a \tan \theta = b \]
Again, substituting for \( \sec \theta \) and \( \tan \theta \):
\[ q \frac{1}{\cos \theta} + a \frac{\sin \theta}{\cos \theta} = b \]
This can be rewritten as:
\[ \frac{q + a \sin \theta}{\cos \theta} = b \]

4. **Multiply both sides by \( \cos \theta \):**
\[ q + a \sin \theta = b \cos \theta \]
Rearranging gives us:
\[ q = b \cos \theta - a \sin \theta \quad \text{(Equation 2)} \]

5. **Now we will square both equations to find \( P^2 + Q^2 \):**
- From Equation 1:
\[ P^2 = (a \cos \theta + b \sin \theta)^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta \]
- From Equation 2:
\[ Q^2 = (b \cos \theta - a \sin \theta)^2 = b^2 \cos^2 \theta + a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta \]

6. **Adding \( P^2 \) and \( Q^2 \):**
\[ P^2 + Q^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta) + (b^2 \cos^2 \theta + a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) \]
Simplifying this:
\[ P^2 + Q^2 = (a^2 + b^2)(\cos^2 \theta + \sin^2 \theta) \]

7. **Using the Pythagorean identity:**
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Therefore:
\[ P^2 + Q^2 = a^2 + b^2 \]

### Final Result:
The relationship between \( P, Q, A, \) and \( B \) is:
\[ P^2 + Q^2 = A^2 + B^2 \]