If ` sin (x+y) +cos (x+y) =1,then (dy)/(dx)=`

Options

A. ` -1`
B. 1
C. 0
D. 2

Answer

To find \(\frac{dy}{dx}\) given the equation \( \sin(x+y) + \cos(x+y) = 1 \), we will differentiate both sides of the equation with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate the equation**:
We start with the equation:
\[ \sin(x+y) + \cos(x+y) = 1 \]
Now, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}[\sin(x+y)] + \frac{d}{dx}[\cos(x+y)] = \frac{d}{dx}[1] \]

2. **Apply the chain rule**:
Using the chain rule for differentiation:
- The derivative of \(\sin(u)\) is \(\cos(u) \cdot \frac{du}{dx}\)
- The derivative of \(\cos(u)\) is \(-\sin(u) \cdot \frac{du}{dx}\)
where \(u = x + y\). Thus, we have:
\[ \cos(x+y) \cdot \left(1 + \frac{dy}{dx}\right) - \sin(x+y) \cdot \left(1 + \frac{dy}{dx}\right) = 0 \]

3. **Factor out the common term**:
We can factor out \((1 + \frac{dy}{dx})\):
\[ (1 + \frac{dy}{dx})(\cos(x+y) - \sin(x+y)) = 0 \]

4. **Set the factors to zero**:
This gives us two cases:
- Case 1: \(1 + \frac{dy}{dx} = 0\)
- Case 2: \(\cos(x+y) - \sin(x+y) = 0\)

5. **Solve Case 1**:
From Case 1:
\[ 1 + \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -1 \]

6. **Solve Case 2**:
From Case 2:
\[ \cos(x+y) = \sin(x+y) \]
This implies that \(x+y = \frac{\pi}{4} + n\pi\) for any integer \(n\). However, this does not provide a value for \(\frac{dy}{dx}\) directly.

7. **Final Result**:
The only relevant solution for \(\frac{dy}{dx}\) is from Case 1:
\[ \frac{dy}{dx} = -1 \]

### Conclusion:
Thus, the final answer is:
\[ \frac{dy}{dx} = -1 \]