If `sin^(-1)x+sin^(-1)y=(2pi)/(3), cos^(-1)x-cos^(-1)y=(pi)/(3)` then the number of values of (x, y) is :

Options

A. two
B. four
C. zero
D. None of these

Answer

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

1. \( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \)
2. \( \cos^{-1} x - \cos^{-1} y = \frac{\pi}{3} \)

### Step 1: Rewrite the equations using known identities

We know that:
- \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \)

Using this identity, we can rewrite the second equation:

\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]
Substituting this into the second equation gives:

\[ \left(\frac{\pi}{2} - \sin^{-1} x\right) - \cos^{-1} y = \frac{\pi}{3} \]

### Step 2: Simplify the second equation

Rearranging the second equation, we get:

\[ \frac{\pi}{2} - \sin^{-1} x - \cos^{-1} y = \frac{\pi}{3} \]

Now, we can express \( \cos^{-1} y \) in terms of \( \sin^{-1} y \):

\[ \cos^{-1} y = \frac{\pi}{2} - \sin^{-1} y \]

Substituting this into the equation gives:

\[ \frac{\pi}{2} - \sin^{-1} x - \left(\frac{\pi}{2} - \sin^{-1} y\right) = \frac{\pi}{3} \]

This simplifies to:

\[ \sin^{-1} y - \sin^{-1} x = \frac{\pi}{3} \]

### Step 3: Set up the equations

Now we have two equations:

1. \( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \)
2. \( \sin^{-1} y - \sin^{-1} x = \frac{\pi}{3} \)

### Step 4: Solve the equations

Let \( a = \sin^{-1} x \) and \( b = \sin^{-1} y \). We can rewrite the equations as:

1. \( a + b = \frac{2\pi}{3} \)
2. \( b - a = \frac{\pi}{3} \)

Now, we can solve these equations simultaneously.

Adding the two equations:

\[ (a + b) + (b - a) = \frac{2\pi}{3} + \frac{\pi}{3} \]

This simplifies to:

\[ 2b = \pi \implies b = \frac{\pi}{2} \]

Substituting \( b = \frac{\pi}{2} \) back into the first equation:

\[ a + \frac{\pi}{2} = \frac{2\pi}{3} \]

This simplifies to:

\[ a = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{4\pi - 3\pi}{6} = \frac{\pi}{6} \]

### Step 5: Find \( x \) and \( y \)

Now we have:

\[ \sin^{-1} x = \frac{\pi}{6} \implies x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

\[ \sin^{-1} y = \frac{\pi}{2} \implies y = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Conclusion

Thus, the ordered pair \( (x, y) \) is \( \left(\frac{1}{2}, 1\right) \).

### Number of Values

Since we derived unique values for \( x \) and \( y \), there is only **1 unique pair** \( (x, y) \).