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. `cos.(pi^(2))/(8)`
B. `sin.(pi^(2))/(4)`
C. `cos.(pi^(2))/(2)`
D. none of these

Explanation

To solve the problem, we need to analyze the given equations step by step.

### Step 1: Write down the equations
We have two equations:
1. \((\sin^{-1} y)^2 + \cos^{-1} x = \frac{n \pi^2}{4}\) (Equation 1)
2. \((\sin^{-1} y)^2 - \cos^{-1} x = \frac{\pi^2}{16}\) (Equation 2)

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

### Step 3: 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 have:
\[ 2(\sin^{-1} y)^2 = \frac{4n \pi^2 + \pi^2}{16} = \frac{(4n + 1) \pi^2}{16} \]

### Step 4: Divide both sides by 2
Dividing both sides by 2 gives:
\[ (\sin^{-1} y)^2 = \frac{(4n + 1) \pi^2}{32} \]

### Step 5: Determine the range of \((\sin^{-1} y)^2\)
Since \(\sin^{-1} y\) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), \((\sin^{-1} y)^2\) ranges from \(0\) to \(\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}\). Thus, we have:
\[ 0 \leq (\sin^{-1} y)^2 < \frac{\pi^2}{4} \]

### Step 6: Set up inequalities
From the equation we derived:
\[ 0 \leq \frac{(4n + 1) \pi^2}{32} < \frac{\pi^2}{4} \]
This leads to two inequalities:
1. \(0 \leq 4n + 1\)
2. \(4n + 1 < 8\)

### Step 7: Solve the inequalities
1. From \(0 \leq 4n + 1\), we get:
\[ 4n \geq -1 \implies n \geq -\frac{1}{4} \]
Since \(n\) is a natural number, this condition is satisfied for \(n \geq 1\).

2. From \(4n + 1 < 8\), we get:
\[ 4n < 7 \implies n < \frac{7}{4} \]
Thus, \(n\) can only take the value \(1\) since it must be a natural number.

### Conclusion
The only natural number \(n\) that satisfies both inequalities is \(n = 1\).

### Final Answer
Thus, \(n\) can be equal to \(1\).

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