Find the value of `acosθ−bsinθ` if `asinθ+bcosθ=c`.

Answer

To find the value of \( a \cos \theta - b \sin \theta \) given that \( a \sin \theta + b \cos \theta = c \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ a \sin \theta + b \cos \theta = c \]

2. **Square both sides of the equation:**
\[ (a \sin \theta + b \cos \theta)^2 = c^2 \]

3. **Expand the left-hand side using the formula \((x + y)^2 = x^2 + y^2 + 2xy\):**
\[ a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta = c^2 \]

4. **Use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):**
- Rewrite \(\sin^2 \theta\) as \(1 - \cos^2 \theta\) and \(\cos^2 \theta\) as \(1 - \sin^2 \theta\).
\[ a^2 (1 - \cos^2 \theta) + b^2 (1 - \sin^2 \theta) + 2ab \sin \theta \cos \theta = c^2 \]

5. **Distribute \(a^2\) and \(b^2\):**
\[ a^2 - a^2 \cos^2 \theta + b^2 - b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta = c^2 \]

6. **Combine like terms:**
\[ (a^2 + b^2) - (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + 2ab \sin \theta \cos \theta = c^2 \]

7. **Rearrange the equation to isolate the terms involving \(\cos \theta\) and \(\sin \theta\):**
\[ a^2 + b^2 - c^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta \]

8. **Recognize the right-hand side as a perfect square:**
\[ a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta = (a \cos \theta - b \sin \theta)^2 \]

9. **Set the two sides equal:**
\[ a^2 + b^2 - c^2 = (a \cos \theta - b \sin \theta)^2 \]

10. **Take the square root of both sides:**
\[ a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} \]

### Final Answer:
\[ a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} \]