If `nin N` and the set of equations, `(sin^-1 y)^2 + (cos^-1 x)=(n pi^2)/4 and (sin^-1y)^2 - (cos^-1 x) = pi^2/16` is consistent,then n can be equal to-

Options

A. 1
B. `-1`
C. `(1)/(2)`
D. none of these

Explanation

To solve the problem, we need to analyze the given equations and find the value of \( n \) that satisfies the conditions. The equations are:

1. \( (\sin^{-1} y)^2 + \cos^{-1} x = \frac{n \pi^2}{4} \)
2. \( (\sin^{-1} y)^2 - \cos^{-1} x = \frac{\pi^2}{16} \)

We will follow these steps:

### Step 1: Add the two equations
Adding both equations, we get:
\[ (\sin^{-1} y)^2 + \cos^{-1} x + (\sin^{-1} y)^2 - \cos^{-1} x = \frac{n \pi^2}{4} + \frac{\pi^2}{16} \]
This simplifies to:
\[ 2(\sin^{-1} y)^2 = \frac{n \pi^2}{4} + \frac{\pi^2}{16} \]

### Step 2: Simplify the right-hand side
To combine the terms on the right-hand side, we need a common denominator:
\[ \frac{n \pi^2}{4} = \frac{4n \pi^2}{16} \]
Thus, we can write:
\[ 2(\sin^{-1} y)^2 = \frac{4n \pi^2 + \pi^2}{16} = \frac{(4n + 1) \pi^2}{16} \]

### Step 3: Isolate \( (\sin^{-1} y)^2 \)
Dividing both sides by 2, we have:
\[ (\sin^{-1} y)^2 = \frac{(4n + 1) \pi^2}{32} \]

### Step 4: Take the square root
Taking the square root of both sides gives:
\[ \sin^{-1} y = \sqrt{\frac{(4n + 1) \pi^2}{32}} = \frac{\pi \sqrt{4n + 1}}{4\sqrt{2}} \]

### Step 5: Determine the range of \( \sin^{-1} y \)
The range of \( \sin^{-1} y \) is from \( 0 \) to \( \frac{\pi}{2} \). Therefore, we have:
\[ 0 \leq \frac{\pi \sqrt{4n + 1}}{4\sqrt{2}} \leq \frac{\pi}{2} \]

### Step 6: Remove \( \pi \) and solve the inequality
Dividing through by \( \pi \) (which is positive), we get:
\[ 0 \leq \frac{\sqrt{4n + 1}}{4\sqrt{2}} \leq \frac{1}{2} \]
Multiplying through by \( 4\sqrt{2} \):
\[ 0 \leq \sqrt{4n + 1} \leq 2\sqrt{2} \]

### Step 7: Square the inequality
Squaring the inequality gives:
\[ 0 \leq 4n + 1 \leq 8 \]
This simplifies to:
\[ -1 \leq 4n \leq 7 \]

### Step 8: Solve for \( n \)
Dividing through by 4:
\[ -\frac{1}{4} \leq n \leq \frac{7}{4} \]

### Step 9: Identify natural numbers
Since \( n \) must be a natural number (i.e., \( n \in \mathbb{N} \)), the only natural number that satisfies this inequality is:
\[ n = 1 \]

### Conclusion
Thus, the value of \( n \) can be equal to:
\[ \boxed{1} \]