If `a cos theta + b sin theta =4 and a sin theta- b cos theta = 3 " then" a^(2) + b^(2)` is equal to

Options

A. 7
B. 12
C. 25
D. none of these

Explanation

To solve the problem, we need to find the value of \( a^2 + b^2 \) given the equations:

1. \( a \cos \theta + b \sin \theta = 4 \)
2. \( a \sin \theta - b \cos \theta = 3 \)

### Step-by-Step Solution:

**Step 1: Square both equations.**

Starting with the first equation:
\[ (a \cos \theta + b \sin \theta)^2 = 4^2 \]
This expands to:
\[ a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta = 16 \tag{1} \]

Now for the second equation:
\[ (a \sin \theta - b \cos \theta)^2 = 3^2 \]
This expands to:
\[ a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = 9 \tag{2} \]

**Step 2: Add the two squared equations.**

Now, we add equations (1) and (2):
\[ (a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \cos \theta \sin \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta) = 16 + 9 \]
This simplifies to:
\[ a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = 25 \]

**Step 3: Use the Pythagorean identity.**

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ a^2 (1) + b^2 (1) = 25 \]
Thus, we have:
\[ a^2 + b^2 = 25 \]

### Final Answer:
\[ \boxed{25} \]