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

Options

A. `sqrt(3)`
B. `pi/3`
C. 1
D. 0

Answer

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

2. **Use the identity:**
We know that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
Therefore, 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 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. **Simplify the equation:**
\[ \frac{\pi}{2} + \frac{\pi}{2} - \cos^{-1}x - \cos^{-1}y = \frac{2\pi}{3} \]
This simplifies to:
\[ \pi - \cos^{-1}x - \cos^{-1}y = \frac{2\pi}{3} \]

5. **Rearrange to isolate \( \cos^{-1}x + \cos^{-1}y \):**
\[ -(\cos^{-1}x + \cos^{-1}y) = \frac{2\pi}{3} - \pi \]
\[ -(\cos^{-1}x + \cos^{-1}y) = \frac{2\pi}{3} - \frac{3\pi}{3} \]
\[ -(\cos^{-1}x + \cos^{-1}y) = -\frac{\pi}{3} \]

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

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