If a `sin theta + b cos theta = c` then the value of a `cos theta -b sin theta` is :

Options

A. `pmsqrt(-a^2 + b^2 + c^2)`
B. `pm sqrt(a^2 + b^2 -c^2)`
C. `pm sqrt(a^2 - b^2 - c^2)`
D. `pm sqrt(a^2 - b^2 + c^2)`

Answer

To solve the problem, we need to find the value of \( a \cos \theta - b \sin \theta \) given that \( a \sin \theta + b \cos \theta = c \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ a \sin \theta + b \cos \theta = c \]
Let's denote this as Equation (1).

2. **Let \( x = a \cos \theta - b \sin \theta \):**
We need to find the value of \( x \). Let's denote this as Equation (2).

3. **Square both equations:**
- Square Equation (1):
\[ (a \sin \theta + b \cos \theta)^2 = c^2 \]
Expanding this gives:
\[ a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = c^2 \]
This is Equation (3).

- Square Equation (2):
\[ (a \cos \theta - b \sin \theta)^2 = x^2 \]
Expanding this gives:
\[ a^2 \cos^2 \theta - 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta = x^2 \]
This is Equation (4).

4. **Add Equations (3) and (4):**
\[ (a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta) + (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) = c^2 + x^2 \]
Simplifying this gives:
\[ a^2 (\sin^2 \theta + \cos^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = c^2 + x^2 \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ a^2 + b^2 = c^2 + x^2 \]

5. **Rearranging the equation:**
\[ x^2 = a^2 + b^2 - c^2 \]

6. **Taking the square root:**
\[ x = \sqrt{a^2 + b^2 - c^2} \quad \text{or} \quad x = -\sqrt{a^2 + b^2 - c^2} \]

### Final Result:
Thus, the value of \( a \cos \theta - b \sin \theta \) is:
\[ x = \pm \sqrt{a^2 + b^2 - c^2} \]