What are the values of (x,y) satisfying the simultaneous equations `sin^(-1)x+sin^(-1)y=(2pi)/(3)` and `cos^(-1)x-cos^(-1)y=(pi)/(3)`?

Options

A. A. (0,1)
B. B. `((1)/(2),1)`
C. C. `(1,(1)/(2))`
D. D. `((sqrt3)/(2),1)`

Answer

To solve the simultaneous equations \( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \) and \( \cos^{-1} x - \cos^{-1} y = -\frac{\pi}{3} \), we will follow these steps:

### Step 1: Write down the equations
The two equations we have are:
1. \( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \) (Equation 1)
2. \( \cos^{-1} x - \cos^{-1} y = -\frac{\pi}{3} \) (Equation 2)

### Step 2: Rearranging Equation 2
From Equation 2, we can rearrange it to:
\[ \cos^{-1} x = \cos^{-1} y - \frac{\pi}{3} \]

### Step 3: Add both equations
Now, we can add Equation 1 and Equation 2:
\[ \sin^{-1} x + \sin^{-1} y + \cos^{-1} x - \cos^{-1} y = \frac{2\pi}{3} - \frac{\pi}{3} \]
This simplifies to:
\[ \sin^{-1} x + \cos^{-1} x + \sin^{-1} y - \cos^{-1} y = \frac{\pi}{3} \]

### Step 4: Use the identity
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Thus, we can substitute this into our equation:
\[ \frac{\pi}{2} + \sin^{-1} y - \cos^{-1} y = \frac{\pi}{3} \]

### Step 5: Isolate \( \sin^{-1} y - \cos^{-1} y \)
Rearranging gives us:
\[ \sin^{-1} y - \cos^{-1} y = \frac{\pi}{3} - \frac{\pi}{2} \]
Calculating the right-hand side:
\[ \frac{\pi}{3} - \frac{\pi}{2} = \frac{2\pi - 3\pi}{6} = -\frac{\pi}{6} \]
So we have:
\[ \sin^{-1} y - \cos^{-1} y = -\frac{\pi}{6} \]

### Step 6: Use the identity for \( y \)
We also know:
\[ \sin^{-1} y + \cos^{-1} y = \frac{\pi}{2} \]
Now we can add these two equations:
1. \( \sin^{-1} y - \cos^{-1} y = -\frac{\pi}{6} \)
2. \( \sin^{-1} y + \cos^{-1} y = \frac{\pi}{2} \)

Adding them gives:
\[ 2 \sin^{-1} y = \frac{\pi}{2} - \frac{\pi}{6} \]
Calculating the right-hand side:
\[ \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]
Thus:
\[ 2 \sin^{-1} y = \frac{\pi}{3} \]
Dividing by 2:
\[ \sin^{-1} y = \frac{\pi}{6} \]

### Step 7: Find \( y \)
Taking the sine of both sides:
\[ y = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Step 8: Substitute \( y \) back to find \( x \)
Now substitute \( y \) back into Equation 1:
\[ \sin^{-1} x + \sin^{-1}\left(\frac{1}{2}\right) = \frac{2\pi}{3} \]
Since \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \):
\[ \sin^{-1} x + \frac{\pi}{6} = \frac{2\pi}{3} \]
Subtracting \( \frac{\pi}{6} \) from both sides:
\[ \sin^{-1} x = \frac{2\pi}{3} - \frac{\pi}{6} \]
Calculating the right-hand side:
\[ \frac{2\pi}{3} - \frac{\pi}{6} = \frac{4\pi - \pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]

### Step 9: Find \( x \)
Taking the sine of both sides:
\[ x = \sin\left(\frac{\pi}{2}\right) = 1 \]

### Final Answer
Thus, the values of \( (x, y) \) are:
\[ (x, y) = \left(1, \frac{1}{2}\right) \]