If `a cos theta + b sin theta = p and a sin theta - b cos theta = q`, then the relation between a,b,p and q is

Options

A. `a^2 - b^2 = p^2 - q^2`
B. `a^2 + b^2 = p^2 + q^2`
C. `a + b = p + q`
D. `a - b = p - q`

Answer

To find the relation between \( a, b, p, \) and \( q \) given the equations \( a \cos \theta + b \sin \theta = p \) and \( a \sin \theta - b \cos \theta = q \), we can follow these steps:

### Step 1: Square both equations
We start by squaring both equations.

1. Square the first equation:
\[ (a \cos \theta + b \sin \theta)^2 = p^2 \]
Expanding this, we get:
\[ a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta = p^2 \]

2. Square the second equation:
\[ (a \sin \theta - b \cos \theta)^2 = q^2 \]
Expanding this, we get:
\[ a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = q^2 \]

### Step 2: Add the two squared equations
Now, we add the results from Step 1:
\[ (a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta) = p^2 + q^2 \]

### Step 3: Simplify the equation
Combining like terms:
- For \( a^2 \):
\[ a^2 \cos^2 \theta + a^2 \sin^2 \theta = a^2 (\cos^2 \theta + \sin^2 \theta) = a^2 \]
- For \( b^2 \):
\[ b^2 \sin^2 \theta + b^2 \cos^2 \theta = b^2 (\sin^2 \theta + \cos^2 \theta) = b^2 \]
- The terms involving \( ab \) cancel out:
\[ 2ab \cos \theta \sin \theta - 2ab \sin \theta \cos \theta = 0 \]

Thus, we have:
\[ a^2 + b^2 = p^2 + q^2 \]

### Final Relation
The relation between \( a, b, p, \) and \( q \) is:
\[ a^2 + b^2 = p^2 + q^2 \]