If `sin^-1 x + sin^-1 y = (2pi)/3,` then `cos^-1 x + cos^-1 y =`

Options

A. `(2pi)/(3)`
B. `(pi)/(3)`
C. `(pi)/(6)`
D. `pi`

Answer

To solve the problem where \( \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \), we need to find \( \cos^{-1} x + \cos^{-1} y \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3} \]

2. **Use the identity for sine and cosine inverses:**
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This means we can express \( \sin^{-1} x \) and \( \sin^{-1} y \) in terms of \( \cos^{-1} x \) and \( \cos^{-1} y \):
\[ \sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x \]
\[ \sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y \]

3. **Substitute these expressions into the original equation:**
\[ \left( \frac{\pi}{2} - \cos^{-1} x \right) + \left( \frac{\pi}{2} - \cos^{-1} y \right) = \frac{2\pi}{3} \]

4. **Combine the terms:**
\[ \pi - \cos^{-1} x - \cos^{-1} y = \frac{2\pi}{3} \]

5. **Rearranging the equation:**
\[ - \cos^{-1} x - \cos^{-1} y = \frac{2\pi}{3} - \pi \]
\[ - \cos^{-1} x - \cos^{-1} y = \frac{2\pi}{3} - \frac{3\pi}{3} \]
\[ - \cos^{-1} x - \cos^{-1} y = -\frac{\pi}{3} \]

6. **Multiply through by -1:**
\[ \cos^{-1} x + \cos^{-1} y = \frac{\pi}{3} \]

### Final Answer:
Thus, the value of \( \cos^{-1} x + \cos^{-1} y \) is:
\[ \cos^{-1} x + \cos^{-1} y = \frac{\pi}{3} \]