If `f(sinx)-f(-sinx)=x^(2)-1` is defined for all ` x in R ` , then the value of `x^(2)-2` can be

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

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

\[ f(\sin x) - f(-\sin x) = x^2 - 1 \]

This equation is defined for all \( x \in \mathbb{R} \). We need to find the value of \( x^2 - 2 \).

### Step 1: Substitute \( -x \) into the equation

We replace \( x \) with \( -x \) in the original equation:

\[ f(\sin(-x)) - f(-\sin(-x)) = (-x)^2 - 1 \]

Since \( \sin(-x) = -\sin x \), this simplifies to:

\[ f(-\sin x) - f(\sin x) = x^2 - 1 \]

### Step 2: Rewrite the new equation

The new equation can be rewritten as:

\[ - (f(\sin x) - f(-\sin x)) = x^2 - 1 \]

This implies:

\[ f(-\sin x) - f(\sin x) = x^2 - 1 \]

### Step 3: Add the two equations

Now we have two equations:

1. \( f(\sin x) - f(-\sin x) = x^2 - 1 \) (Equation 1)
2. \( f(-\sin x) - f(\sin x) = x^2 - 1 \) (Equation 2)

If we add these two equations, we get:

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

This simplifies to:

\[ 0 = 2(x^2 - 1) \]

### Step 4: Solve for \( x^2 \)

From the equation \( 0 = 2(x^2 - 1) \), we can divide both sides by 2 (since 2 is not zero):

\[ 0 = x^2 - 1 \]

This gives us:

\[ x^2 = 1 \]

### Step 5: Find \( x^2 - 2 \)

Now we need to find the value of \( x^2 - 2 \):

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

### Conclusion

Thus, the value of \( x^2 - 2 \) can be:

\[ \boxed{-1} \]