If `sin^(-1)x+sin^(-1)y=pi//2` and `cos^(-1)x-cos^(-1)y=0`, then values x and y are respectively

Options

A. `(1)/(sqrt2),-(1)/(sqrt2)`
B. `(1)/(2),(1)/(2)`
C. `(1)/(2),-(1)/(2)`
D. `(1)/(sqrt2).(1)/(sqrt2)`

Answer

To solve the problem, we are given two equations involving inverse trigonometric functions:

1. \( \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \) (Equation 1)
2. \( \cos^{-1} x - \cos^{-1} y = 0 \) (Equation 2)

We need to find the values of \( x \) and \( y \).

### Step 1: Rewrite Equation 2
From Equation 2, we can rewrite it as:
\[ \cos^{-1} x = \cos^{-1} y \]
This implies:
\[ x = y \quad \text{(since the cosine function is one-to-one in the range [0, π])} \]

### Step 2: Substitute \( y \) in Equation 1
Now, substituting \( y \) with \( x \) in Equation 1:
\[ \sin^{-1} x + \sin^{-1} x = \frac{\pi}{2} \]
This simplifies to:
\[ 2 \sin^{-1} x = \frac{\pi}{2} \]

### Step 3: Solve for \( \sin^{-1} x \)
Dividing both sides by 2:
\[ \sin^{-1} x = \frac{\pi}{4} \]

### Step 4: Find \( x \)
Taking the sine of both sides:
\[ x = \sin\left(\frac{\pi}{4}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus,
\[ x = \frac{1}{\sqrt{2}} \]

### Step 5: Find \( y \)
Since we established that \( y = x \), we have:
\[ y = \frac{1}{\sqrt{2}} \]

### Conclusion
The values of \( x \) and \( y \) are:
\[ x = \frac{1}{\sqrt{2}}, \quad y = \frac{1}{\sqrt{2}} \]