If `sin ^(-1) x + sin ^(-1) y = (pi)/(6)`, then `cos^(-1) x + cos ^(-1) y =?`

Options

A. `(pi)/(6)`
B. `(5pi)/(6)`
C. `(pi)/(3)`
D. `(2pi)/(3)`

Answer

To solve the problem, we start with the equation given:

\[ \sin^{-1} x + \sin^{-1} y = \frac{\pi}{6} \]

We need to find the value of:

\[ \cos^{-1} x + \cos^{-1} y \]

### Step 1: Use the identity for inverse sine and cosine
We know the identity:

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

Using this identity, we can express \(\sin^{-1} x\) and \(\sin^{-1} y\) in terms of \(\cos^{-1} x\) and \(\cos^{-1} y\):

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

### Step 2: Substitute into the original equation
Substituting these expressions into the original equation gives:

\[ \left(\frac{\pi}{2} - \cos^{-1} x\right) + \left(\frac{\pi}{2} - \cos^{-1} y\right) = \frac{\pi}{6} \]

### Step 3: Simplify the equation
Now, we simplify the left-hand side:

\[ \frac{\pi}{2} + \frac{\pi}{2} - \cos^{-1} x - \cos^{-1} y = \frac{\pi}{6} \]

This simplifies to:

\[ \pi - (\cos^{-1} x + \cos^{-1} y) = \frac{\pi}{6} \]

### Step 4: Isolate \(\cos^{-1} x + \cos^{-1} y\)
Now, we isolate \(\cos^{-1} x + \cos^{-1} y\):

\[ \cos^{-1} x + \cos^{-1} y = \pi - \frac{\pi}{6} \]

### Step 5: Calculate the right-hand side
To calculate \(\pi - \frac{\pi}{6}\), we convert \(\pi\) into a fraction with a denominator of 6:

\[ \pi = \frac{6\pi}{6} \]

Thus,

\[ \cos^{-1} x + \cos^{-1} y = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6} \]

### Final Answer
So, the value of \(\cos^{-1} x + \cos^{-1} y\) is:

\[ \boxed{\frac{5\pi}{6}} \]