If `sin^(-1)x+sin^(-1)y=pi/2,t h e n(1+x^4+y^4)/(x^2-x^2y^2+y^2)` is equal to 1 (b) 2 (c) `1/2` (d) none of these

Options

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

Explanation

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

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

### Step 1: Rewrite the equation
From the identity \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), we can deduce that:

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

### Step 2: Convert to sine terms
Using the cosine inverse identity, we can express \( y \) in terms of \( x \):

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

### Step 3: Square both sides
Squaring both sides gives us:

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

### Step 4: Rearranging the equation
Rearranging the equation results in:

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

### Step 5: Substitute into the expression
Now we need to evaluate the expression:

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

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

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

### Step 7: Substitute back into the expression
Now substituting back into the expression:

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

### Step 8: Factor out the numerator
Factoring out 2 from the numerator gives:

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

### Step 9: 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 10: Final simplification
Now substituting this back into our expression results in:

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

Thus, the final answer is:

\[ \boxed{2} \]