If ` tan ^(-1) x + tan ^(-1) .sqrt( 1 - y^(2))/y = pi/3 " and " sin^(-1) y - cos^(-1) ( x/(sqrt( 1 + x^(2)))) = pi/6 " , then " ( 5 sin^(-1) x)/( sin^(-1) y) ` is

Answer

To solve the problem, we need to analyze the two equations given and find the value of \( \frac{5 \sin^{-1} x}{\sin^{-1} y} \).

### Step 1: Analyze the first equation
The first equation is:
\[ \tan^{-1} x + \tan^{-1} \left( \frac{\sqrt{1 - y^2}}{y} \right) = \frac{\pi}{3} \]
Let \( x = \tan a \) and \( y = \cos b \). Then we can rewrite the equation as:
\[ \tan^{-1}(\tan a) + \tan^{-1}(\tan b) = \frac{\pi}{3} \]
This simplifies to:
\[ a + b = \frac{\pi}{3} \tag{1} \]

### Step 2: Analyze the second equation
The second equation is:
\[ \sin^{-1} y - \cos^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right) = \frac{\pi}{6} \]
Using \( y = \cos b \) and \( x = \tan a \), we can rewrite this as:
\[ \sin^{-1}(\cos b) - \cos^{-1}(\sin a) = \frac{\pi}{6} \]
Using the identity \( \sin^{-1}(\cos b) = \frac{\pi}{2} - b \), the equation becomes:
\[ \frac{\pi}{2} - b - \left( \frac{\pi}{2} - a \right) = \frac{\pi}{6} \]
This simplifies to:
\[ a - b = \frac{\pi}{6} \tag{2} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( a + b = \frac{\pi}{3} \)
2. \( a - b = \frac{\pi}{6} \)

Adding these two equations:
\[ (a + b) + (a - b) = \frac{\pi}{3} + \frac{\pi}{6} \]
\[ 2a = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]
Thus, we find:
\[ a = \frac{\pi}{4} \]

Now substituting \( a \) back into equation (1):
\[ \frac{\pi}{4} + b = \frac{\pi}{3} \]
\[ b = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

### Step 4: Find \( x \) and \( y \)
Now we can find \( x \) and \( y \):
\[ x = \tan a = \tan\left(\frac{\pi}{4}\right) = 1 \]
\[ y = \cos b = \cos\left(\frac{\pi}{12}\right) \]

### Step 5: Calculate \( \frac{5 \sin^{-1} x}{\sin^{-1} y} \)
Now we need to find:
\[ \frac{5 \sin^{-1} x}{\sin^{-1} y} \]
Since \( x = 1 \):
\[ \sin^{-1} 1 = \frac{\pi}{2} \]
For \( y = \cos\left(\frac{\pi}{12}\right) \):
\[ \sin^{-1} y = \sin^{-1}(\cos(\frac{\pi}{12})) = \frac{\pi}{2} - \frac{\pi}{12} = \frac{6\pi - \pi}{12} = \frac{5\pi}{12} \]

Thus:
\[ \frac{5 \sin^{-1} x}{\sin^{-1} y} = \frac{5 \cdot \frac{\pi}{2}}{\frac{5\pi}{12}} = \frac{5 \cdot \frac{\pi}{2} \cdot 12}{5\pi} = \frac{12}{2} = 6 \]

### Final Answer
The value of \( \frac{5 \sin^{-1} x}{\sin^{-1} y} \) is \( 6 \).