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: Set up the equations
Given:
1. \( \sin x - \cos x = 1 \) (Equation 1)
2. Let \( \sin x + \cos x = y \) (Equation 2)

### Step 2: Square both equations
Square Equation 1:
\[ (\sin x - \cos x)^2 = 1^2 \]
Expanding this:
\[ \sin^2 x - 2 \sin x \cos x + \cos^2 x = 1 \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 1 - 2 \sin x \cos x = 1 \]
This simplifies to:
\[ -2 \sin x \cos x = 0 \quad \text{(Equation 3)} \]

### Step 3: Square Equation 2
Square Equation 2:
\[ (\sin x + \cos x)^2 = y^2 \]
Expanding this:
\[ \sin^2 x + 2 \sin x \cos x + \cos^2 x = y^2 \]
Again using the Pythagorean identity:
\[ 1 + 2 \sin x \cos x = y^2 \quad \text{(Equation 4)} \]

### Step 4: Add Equation 3 and Equation 4
From Equation 3, we have:
\[ -2 \sin x \cos x = 0 \implies \sin x \cos x = 0 \]
This means either \( \sin x = 0 \) or \( \cos x = 0 \). However, since \( x \) is an acute angle, neither can be true.

Now, substitute \( \sin x \cos x = 0 \) into Equation 4:
\[ 1 + 2(0) = y^2 \]
Thus:
\[ y^2 = 1 \]

### Step 5: Solve for \( y \)
Taking the square root of both sides:
\[ y = 1 \quad \text{(since \( y \) must be positive for acute angles)} \]

### Conclusion
Thus, the value of \( \sin x + \cos x \) is:
\[ \sin x + \cos x = 1 \]

### Final Answer
The value of \( \sin x + \cos x \) is \( 1 \).

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