If `"sin"^(-1)x+"sin"^(-1)y=pi/2`, then `(1+x^4+y^4)`/(`x^2-x^2y^2+y^2)` is equal to

Options

A. 1
B. 2
C. `1//2`
D. None of these

Answer

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

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

### Step 1: Rewrite the equation using the cosine inverse function
From the identity, we know that:

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

### Step 2: Use the right triangle definition
Let \(\alpha = \cos^{-1}y\). In a right triangle, we can express \(x\) and \(y\) in terms of the sides of the triangle:

- The adjacent side to angle \(\alpha\) is \(y\).
- The hypotenuse is \(1\).
- The opposite side can be calculated using Pythagoras' theorem:

\[ \text{Opposite side} = \sqrt{1 - y^2} \]

Thus, we can express \(x\) as:

\[ x = \sin(\alpha) = \frac{\sqrt{1 - y^2}}{1} = \sqrt{1 - y^2} \]

### Step 3: Establish the relationship between \(x\) and \(y\)
From the above, we have:

\[ x^2 = 1 - y^2 \]

### Step 4: Find \(x^2 + y^2\)
Adding \(x^2\) and \(y^2\):

\[ x^2 + y^2 = (1 - y^2) + y^2 = 1 \]

### Step 5: Substitute into the expression
We need to evaluate:

\[ \frac{1 + x^4 + y^4}{x^2 - x^2y^2 + y^2} \]

### Step 6: Simplify the numerator
Using the identity \(x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2\):

\[ x^4 + y^4 = 1^2 - 2x^2y^2 = 1 - 2x^2y^2 \]

Thus, the numerator becomes:

\[ 1 + x^4 + y^4 = 1 + (1 - 2x^2y^2) = 2 - 2x^2y^2 \]

### Step 7: Simplify the denominator
The denominator can be rewritten as:

\[ x^2 - x^2y^2 + y^2 = (x^2 + y^2) - x^2y^2 = 1 - x^2y^2 \]

### Step 8: Combine the results
Now we can substitute back into our expression:

\[ \frac{2 - 2x^2y^2}{1 - x^2y^2} \]

### Step 9: Factor out common terms
Factoring out \(2\) from the numerator:

\[ = \frac{2(1 - x^2y^2)}{1 - x^2y^2} = 2 \]

### Final Result
Thus, the value of the expression is:

\[ \boxed{2} \]