Solution set of inequation
`(cos^(-1)x)^(2)-(sin^(-1)x)^(2)gt0` is

Options

A. `[0, (1)/(sqrt(2)))`
B. `[-1, (1)/(sqrt(2)))`
C. `(-1, 1)`
D. `[-1, (1)/(2))`

Answer

To solve the inequality \((\cos^{-1} x)^2 - (\sin^{-1} x)^2 > 0\), we can follow these steps:

### Step 1: Rewrite the expression
We recognize that the expression is in the form of \(a^2 - b^2\), which can be factored using the identity \(a^2 - b^2 = (a + b)(a - b)\). Here, let \(a = \cos^{-1} x\) and \(b = \sin^{-1} x\).

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

### Step 2: Simplify the first factor
We know that \(\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}\). Therefore, we can substitute this into our expression:

\[ \left(\frac{\pi}{2}\right)(\cos^{-1} x - \sin^{-1} x) > 0 \]

### Step 3: Analyze the inequality
Since \(\frac{\pi}{2}\) is a positive constant, we can simplify the inequality to:

\[ \cos^{-1} x - \sin^{-1} x > 0 \]

### Step 4: Rewrite \(\cos^{-1} x\)
Using the identity \(\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x\), we substitute:

\[ \frac{\pi}{2} - \sin^{-1} x - \sin^{-1} x > 0 \]

This simplifies to:

\[ \frac{\pi}{2} - 2\sin^{-1} x > 0 \]

### Step 5: Rearranging the inequality
Rearranging gives us:

\[ 2\sin^{-1} x < \frac{\pi}{2} \]

### Step 6: Divide by 2
Dividing both sides by 2 results in:

\[ \sin^{-1} x < \frac{\pi}{4} \]

### Step 7: Apply the sine function
Now, applying the sine function to both sides, we have:

\[ x < \sin\left(\frac{\pi}{4}\right) \]

Since \(\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\), we conclude:

\[ x < \frac{1}{\sqrt{2}} \]

### Step 8: Determine the range of \(x\)
The domain of \(\sin^{-1} x\) is \([-1, 1]\). Therefore, combining this with our previous result, we find:

\[ -1 \leq x < \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the solution set of the inequality is:

\[ x \in [-1, \frac{1}{\sqrt{2}}) \]