If `sin(x-y)=cos(x+y)=1/2,` the value of x and y lyingbetween `0^(@)and 90^(@)` are given by

Options

A. `x=15^(@),y=25^(@)`
B. `x=65^(@),y=15^(@)`
C. `x=45^(@),y=45^(@)`
D. `x=45^(@),y=15^(@)`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) given that:

\[ \sin(x - y) = \frac{1}{2} \quad \text{and} \quad \cos(x + y) = \frac{1}{2} \]

### Step 1: Solve for \( x - y \)

From the equation \( \sin(x - y) = \frac{1}{2} \), we know that:

\[ x - y = 30^\circ \quad \text{or} \quad x - y = 150^\circ \]

However, since \( x \) and \( y \) are both between \( 0^\circ \) and \( 90^\circ \), we will only consider:

\[ x - y = 30^\circ \quad \text{(Equation 1)} \]

### Step 2: Solve for \( x + y \)

From the equation \( \cos(x + y) = \frac{1}{2} \), we know that:

\[ x + y = 60^\circ \quad \text{or} \quad x + y = 300^\circ \]

Again, since \( x \) and \( y \) are both between \( 0^\circ \) and \( 90^\circ \), we will only consider:

\[ x + y = 60^\circ \quad \text{(Equation 2)} \]

### Step 3: Add Equation 1 and Equation 2

Now, we can add Equation 1 and Equation 2:

\[ (x - y) + (x + y) = 30^\circ + 60^\circ \]

This simplifies to:

\[ 2x = 90^\circ \]

Dividing both sides by 2 gives:

\[ x = 45^\circ \]

### Step 4: Substitute \( x \) back to find \( y \)

Now, we can substitute \( x = 45^\circ \) back into Equation 1 to find \( y \):

\[ 45^\circ - y = 30^\circ \]

Solving for \( y \):

\[ y = 45^\circ - 30^\circ = 15^\circ \]

### Conclusion

Thus, the values of \( x \) and \( y \) are:

\[ x = 45^\circ, \quad y = 15^\circ \]

### Final Answer

The correct option is \( x = 45^\circ \) and \( y = 15^\circ \).

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