The set of value of x, satisfying the equation `tan^(2)(sin^(-1)x) gt 1` is :

Options

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

Answer

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

### Step 1: Understanding the Inequality
We start with the inequality:
\[ \tan^2(\sin^{-1} x) > 1 \]
This implies that:
\[ \tan(\sin^{-1} x) > 1 \quad \text{or} \quad \tan(\sin^{-1} x) < -1 \]

### Step 2: Expressing \(\tan(\sin^{-1} x)\)
Using the identity for tangent in terms of sine, we know:
\[ \tan(\sin^{-1} x) = \frac{\sin(\sin^{-1} x)}{\cos(\sin^{-1} x)} = \frac{x}{\sqrt{1 - x^2}} \]
Thus, we can rewrite the inequality as:
\[ \frac{x}{\sqrt{1 - x^2}} > 1 \quad \text{or} \quad \frac{x}{\sqrt{1 - x^2}} < -1 \]

### Step 3: Solving the First Inequality
Let's solve the first part:
\[ \frac{x}{\sqrt{1 - x^2}} > 1 \]
Cross-multiplying (valid since \(\sqrt{1 - x^2} > 0\) for \(|x| < 1\)):
\[ x > \sqrt{1 - x^2} \]
Squaring both sides:
\[ x^2 > 1 - x^2 \]
This simplifies to:
\[ 2x^2 > 1 \quad \Rightarrow \quad x^2 > \frac{1}{2} \quad \Rightarrow \quad |x| > \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ x < -\frac{1}{\sqrt{2}} \quad \text{or} \quad x > \frac{1}{\sqrt{2}} \]

### Step 4: Solving the Second Inequality
Now, we solve the second part:
\[ \frac{x}{\sqrt{1 - x^2}} < -1 \]
Cross-multiplying (valid since \(\sqrt{1 - x^2} > 0\)):
\[ x < -\sqrt{1 - x^2} \]
Squaring both sides:
\[ x^2 < 1 - x^2 \]
This simplifies to:
\[ 2x^2 < 1 \quad \Rightarrow \quad x^2 < \frac{1}{2} \quad \Rightarrow \quad |x| < \frac{1}{\sqrt{2}} \]
Thus, we have:
\[ -\frac{1}{\sqrt{2}} < x < 0 \]

### Step 5: Combining the Results
From both parts, we combine the results:
1. From \(x > \frac{1}{\sqrt{2}}\) or \(x < -\frac{1}{\sqrt{2}}\)
2. From \( -\frac{1}{\sqrt{2}} < x < 0\)

The combined solution set is:
\[ x \in \left(-1, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, 1\right) \]

### Final Answer
Thus, the set of values of \(x\) satisfying the equation \( \tan^2(\sin^{-1} x) > 1 \) is:
\[ x \in \left(-1, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, 1\right) \]