If `acostheta +bsin theta=pand 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:

1. \( a \cos \theta + b \sin \theta = p \)
2. \( a \sin \theta - b \cos \theta = q \)

we will follow these steps:

### Step 1: Square both equations

We start by squaring both equations:

\[ (a \cos \theta + b \sin \theta)^2 = p^2 \]
\[ (a \sin \theta - b \cos \theta)^2 = q^2 \]

### Step 2: Expand both squared equations

Expanding the first equation:

\[ (a \cos \theta + b \sin \theta)^2 = a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta = p^2 \]

Expanding the second equation:

\[ (a \sin \theta - b \cos \theta)^2 = a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = q^2 \]

### Step 3: Add the two expanded equations

Now we add the two expanded equations:

\[ (a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta) + (a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta) = p^2 + q^2 \]

### Step 4: Simplify the equation

Notice that the \( 2ab \cos \theta \sin \theta \) and \( -2ab \sin \theta \cos \theta \) terms will cancel each other out. Thus, we have:

\[ a^2 \cos^2 \theta + a^2 \sin^2 \theta + b^2 \sin^2 \theta + b^2 \cos^2 \theta = p^2 + q^2 \]

### Step 5: Factor out common terms

We can factor out \( a^2 \) and \( b^2 \):

\[ a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = p^2 + q^2 \]

### Step 6: Use the Pythagorean identity

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ a^2 (1) + b^2 (1) = p^2 + q^2 \]

Thus, we arrive at the final relation:

\[ a^2 + b^2 = p^2 + q^2 \]

### Conclusion

The relation between \( a \), \( b \), \( p \), and \( q \) is:

\[ a^2 + b^2 = p^2 + q^2 \]