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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Use the identity for inverse sine and cosine:**
We know that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
and similarly for \( y \):
\[ \sin^{-1}y + \cos^{-1}y = \frac{\pi}{2} \]

2. **Express \( \cos^{-1}x \) and \( \cos^{-1}y \):**
From the above identities, we can express \( \cos^{-1}x \) and \( \cos^{-1}y \) in terms of \( \sin^{-1}x \) and \( \sin^{-1}y \):
\[ \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \]
\[ \cos^{-1}y = \frac{\pi}{2} - \sin^{-1}y \]

3. **Substitute into the equation:**
Now, we substitute these expressions into \( \cos^{-1}x + \cos^{-1}y \):
\[ \cos^{-1}x + \cos^{-1}y = \left(\frac{\pi}{2} - \sin^{-1}x\right) + \left(\frac{\pi}{2} - \sin^{-1}y\right) \]
Simplifying this gives:
\[ \cos^{-1}x + \cos^{-1}y = \pi - (\sin^{-1}x + \sin^{-1}y) \]

4. **Substitute the given value:**
We know from the problem that \( \sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} \). Therefore, we substitute this into our equation:
\[ \cos^{-1}x + \cos^{-1}y = \pi - \frac{2\pi}{3} \]

5. **Perform the subtraction:**
Now, we simplify \( \pi - \frac{2\pi}{3} \):
\[ \pi = \frac{3\pi}{3} \quad \text{(to have a common denominator)} \]
Thus:
\[ \cos^{-1}x + \cos^{-1}y = \frac{3\pi}{3} - \frac{2\pi}{3} = \frac{\pi}{3} \]

### Final Answer:
\[ \cos^{-1}x + \cos^{-1}y = \frac{\pi}{3} \]