If `p = a sin x +b cos x and q = a cos x - b sin x`, then what is the value of `p^2 + q^2` ?

Options

A. a +b
B. ab
C. `a^2 +b^2`
D. `a^2 -b^2`

Answer

To solve the problem, we need to find the value of \( p^2 + q^2 \) given:

\[ p = a \sin x + b \cos x \]
\[ q = a \cos x - b \sin x \]

### Step 1: Calculate \( p^2 \)

First, we find \( p^2 \):

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

Expanding this using the formula \( (A + B)^2 = A^2 + 2AB + B^2 \):

\[ p^2 = a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x \]

### Step 2: Calculate \( q^2 \)

Next, we calculate \( q^2 \):

\[ q^2 = (a \cos x - b \sin x)^2 \]

Again, using the expansion formula:

\[ q^2 = a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x \]

### Step 3: Add \( p^2 \) and \( q^2 \)

Now, we add \( p^2 \) and \( q^2 \):

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

### Step 4: Simplify the expression

Combining like terms:

\[ p^2 + q^2 = a^2 \sin^2 x + b^2 \cos^2 x + a^2 \cos^2 x + b^2 \sin^2 x + 2ab \sin x \cos x - 2ab \sin x \cos x \]

The \( 2ab \sin x \cos x \) terms cancel out:

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

Factoring out \( a^2 \) and \( b^2 \):

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

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

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

Thus, we have:

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

### Final Answer

The value of \( p^2 + q^2 \) is:

\[ \boxed{a^2 + b^2} \]