If sinx-cosx=1,where'x'is an acute angle,the value of (sinx+cosx) is:

Options

A. 0
B. 1
C. `(1)/(2)`
D. 2

Answer

To solve the equation \( \sin x - \cos x = 1 \) and find the value of \( \sin x + \cos x \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ \sin x - \cos x = 1 \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ \sin x = \cos x + 1 \]

### Step 3: Square both sides
Now, we will square both sides of the equation:
\[ (\sin x)^2 = (\cos x + 1)^2 \]
This expands to:
\[ \sin^2 x = \cos^2 x + 2\cos x + 1 \]

### Step 4: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \sin^2 x + \cos^2 x = 1 \]
Substituting \( \sin^2 x \) from the identity into our equation gives:
\[ 1 - \cos^2 x = \cos^2 x + 2\cos x + 1 \]

### Step 5: Rearrange the equation
Rearranging this results in:
\[ 1 - \cos^2 x - 1 = 2\cos x + \cos^2 x \]
This simplifies to:
\[ -\cos^2 x = 2\cos x \]
or
\[ \cos^2 x + 2\cos x = 0 \]

### Step 6: Factor the equation
Factoring gives us:
\[ \cos x (\cos x + 2) = 0 \]
This implies:
\[ \cos x = 0 \quad \text{or} \quad \cos x + 2 = 0 \]
Since \( \cos x + 2 = 0 \) is not possible for acute angles (as cosine cannot be negative), we have:
\[ \cos x = 0 \]

### Step 7: Find \( \sin x \)
If \( \cos x = 0 \), then:
\[ \sin x = 1 \]
This corresponds to \( x = 90^\circ \), which is an acute angle.

### Step 8: Calculate \( \sin x + \cos x \)
Now we can find:
\[ \sin x + \cos x = 1 + 0 = 1 \]

### Final Answer
Thus, the value of \( \sin x + \cos x \) is:
\[ \boxed{1} \]