At what angle must the two forces `(x+y)` and `(x-y)` act so that the resultant may be `sqrt((x^(2)+y^(2)))` :-

Options

A. `"cos"^(-1)(-(x^(2)+y^(2)))/(2(x^(2)-y^(2)))`
B. `"cos"^(-1)(-2(x^(2)-y^(2)))/(x^(2)+y^(2))`
C. `"cos"^(-1)(-(x^(2)+y^(2)))/(x^(2)-y^(2))`
D. `"cos"^(-1)((x^(2)-y^(2)))/(x^(2)+y^(2))`

Explanation

To solve the problem, we need to find the angle θ between the two forces \( F_1 = (x + y) \) and \( F_2 = (x - y) \) such that their resultant is \( R = \sqrt{x^2 + y^2} \).

### Step-by-Step Solution:

1. **Identify the Forces and Resultant**:
- Let \( F_1 = x + y \)
- Let \( F_2 = x - y \)
- The resultant \( R \) is given as \( R = \sqrt{x^2 + y^2} \).

2. **Use the Formula for Resultant of Two Forces**:
The formula for the resultant \( R \) of two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \) is:
\[ R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta} \]

3. **Substituting the Values**:
Substitute \( F_1 \) and \( F_2 \) into the formula:
\[ R = \sqrt{(x + y)^2 + (x - y)^2 + 2(x + y)(x - y) \cos \theta} \]

4. **Expand the Squares**:
- Expand \( (x + y)^2 \):
\[ (x + y)^2 = x^2 + 2xy + y^2 \]
- Expand \( (x - y)^2 \):
\[ (x - y)^2 = x^2 - 2xy + y^2 \]
- Therefore:
\[ (x + y)^2 + (x - y)^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 2x^2 + 2y^2 \]

5. **Substituting Back into the Resultant Formula**:
Now substitute back into the resultant formula:
\[ R = \sqrt{2x^2 + 2y^2 + 2(x^2 - y^2) \cos \theta} \]

6. **Set the Resultant Equal to Given Value**:
Set this equal to the given resultant:
\[ \sqrt{x^2 + y^2} = \sqrt{2(x^2 + y^2) + 2(x^2 - y^2) \cos \theta} \]

7. **Square Both Sides**:
Square both sides to eliminate the square root:
\[ x^2 + y^2 = 2(x^2 + y^2) + 2(x^2 - y^2) \cos \theta \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ x^2 + y^2 - 2(x^2 + y^2) = 2(x^2 - y^2) \cos \theta \]
\[ -x^2 - y^2 = 2(x^2 - y^2) \cos \theta \]

9. **Solving for Cosine**:
\[ \cos \theta = \frac{-x^2 - y^2}{2(x^2 - y^2)} \]

10. **Finding the Angle**:
Finally, the angle \( \theta \) can be found using:
\[ \theta = \cos^{-1} \left( \frac{-x^2 - y^2}{2(x^2 - y^2)} \right) \]