The set of values of x, satisfying the equation `tan^2(sin^-1x) > 1` is -

Options

A. `[-1,1]`
B. `[- sqrt2/2, sqrt2/2]`
C. `(-1,1) - [- sqrt2/2,sqrt2/2]`
D. `[-1,1] - (- sqrt2/2, sqrt2/2)`

Answer

To solve the inequality \( \tan^2(\sin^{-1} x) > 1 \), we will follow a systematic approach.

### Step 1: Understand the Range of \( x \)

The function \( \sin^{-1} x \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, we start with the condition:

\[ -1 \leq x \leq 1 \]

### Step 2: Rewrite the Inequality

The inequality \( \tan^2(\sin^{-1} x) > 1 \) can be rewritten as:

\[ \tan(\sin^{-1} x) > 1 \quad \text{or} \quad \tan(\sin^{-1} x) < -1 \]

### Step 3: Analyze the First Case

**Case 1:** \( \tan(\sin^{-1} x) > 1 \)

We know that:

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

Thus, the inequality becomes:

\[ \frac{x}{\sqrt{1 - x^2}} > 1 \]

### Step 4: Solve the First Case

To solve this inequality, we can cross-multiply (keeping in mind that \( \sqrt{1 - x^2} > 0 \) for \( x \) in \((-1, 1)\)):

\[ x > \sqrt{1 - x^2} \]

Squaring both sides gives:

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

Combining terms leads to:

\[ 2x^2 > 1 \quad \Rightarrow \quad x^2 > \frac{1}{2} \quad \Rightarrow \quad |x| > \frac{1}{\sqrt{2}} \]

This implies:

\[ x > \frac{1}{\sqrt{2}} \quad \text{or} \quad x < -\frac{1}{\sqrt{2}} \]

### Step 5: Analyze the Second Case

**Case 2:** \( \tan(\sin^{-1} x) < -1 \)

Using the same expression for \( \tan(\sin^{-1} x) \):

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

Cross-multiplying gives:

\[ x < -\sqrt{1 - x^2} \]

Squaring both sides results in:

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

Combining terms leads to:

\[ 2x^2 < 1 \quad \Rightarrow \quad x^2 < \frac{1}{2} \quad \Rightarrow \quad |x| < \frac{1}{\sqrt{2}} \]

This implies:

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

### Step 6: Combine Results

From Case 1, we have:

\[ x > \frac{1}{\sqrt{2}} \quad \text{or} \quad x < -\frac{1}{\sqrt{2}} \]

From Case 2, we have:

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

Combining these results, we find the set of values for \( x \):

\[ x \in \left(-1, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, 1\right) \]

### Final Answer

The set of values of \( x \) satisfying the inequality \( \tan^2(\sin^{-1} x) > 1 \) is:

\[ x \in \left(-1, -\frac{1}{\sqrt{2}}\right) \cup \left(\frac{1}{\sqrt{2}}, 1\right) \]