If `sintheta+costheta=x` then `sintheta-costheta=?`

Options

A. `pmsqrt(2-x^(2))`
B. `pmsqrt(2+x^(2))`
C. `pmsqrt(4-x^(2))`
D. `pmsqrt(4+x^(2))`

Answer

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

### Step-by-Step Solution:

1. **Given Equation**:
\[ \sin \theta + \cos \theta = x \]

2. **Square Both Sides**:
\[ (\sin \theta + \cos \theta)^2 = x^2 \]

3. **Expand the Left Side**:
Using the identity \( (A + B)^2 = A^2 + B^2 + 2AB \):
\[ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = x^2 \]

4. **Use the Pythagorean Identity**:
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 + 2 \sin \theta \cos \theta = x^2 \]

5. **Rearranging the Equation**:
\[ 2 \sin \theta \cos \theta = x^2 - 1 \]

6. **Let \( y = \sin \theta - \cos \theta \)**:
Now, we will square \( y \):
\[ (\sin \theta - \cos \theta)^2 = y^2 \]

7. **Expand Using the Identity**:
Using the identity \( (A - B)^2 = A^2 + B^2 - 2AB \):
\[ \sin^2 \theta + \cos^2 \theta - 2 \sin \theta \cos \theta = y^2 \]

8. **Substituting the Pythagorean Identity**:
Again, using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 - 2 \sin \theta \cos \theta = y^2 \]

9. **Substituting the Value of \( 2 \sin \theta \cos \theta \)**:
From step 5, we have \( 2 \sin \theta \cos \theta = x^2 - 1 \):
\[ 1 - (x^2 - 1) = y^2 \]

10. **Simplifying the Equation**:
\[ 1 - x^2 + 1 = y^2 \]
\[ 2 - x^2 = y^2 \]

11. **Taking the Square Root**:
\[ y = \pm \sqrt{2 - x^2} \]

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