For some `x in [-1, 1]` we have `cos^(-1)x =tan^(-1)x`.
The value of `x^(2) ` is :

Options

A. `(sqrt(5)-1)/(2)`
B. `(sqrt(5)+1)/(2)`
C. `(sqrt(3)-1)/(2)`
D. `(sqrt(3)+1)/(2)`

Answer

To solve the equation \( \cos^{-1}(x) = \tan^{-1}(x) \) for \( x \) in the interval \([-1, 1]\), we will follow these steps:

### Step 1: Set up the equation
Let \( \tan^{-1}(x) = a \). Then, we have:
\[ \tan(a) = x \]

### Step 2: Construct a right triangle
From the definition of tangent, we can construct a right triangle where:
- The opposite side (perpendicular) is \( x \)
- The adjacent side (base) is \( 1 \)

Using the Pythagorean theorem, the hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \]

### Step 3: Find \( \cos(a) \)
The cosine of angle \( a \) is given by:
\[ \cos(a) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 4: Substitute into the original equation
Since we have \( \cos^{-1}(x) = a \), we can write:
\[ \cos^{-1}(x) = \cos^{-1}\left(\frac{1}{\sqrt{x^2 + 1}}\right) \]
This implies:
\[ x = \frac{1}{\sqrt{x^2 + 1}} \]

### Step 5: Square both sides
Squaring both sides of the equation gives:
\[ x^2 = \frac{1}{x^2 + 1} \]

### Step 6: Rearrange the equation
Multiplying both sides by \( x^2 + 1 \) leads to:
\[ x^2(x^2 + 1) = 1 \]
This simplifies to:
\[ x^4 + x^2 - 1 = 0 \]

### Step 7: Solve the quadratic equation
Let \( y = x^2 \). The equation becomes:
\[ y^2 + y - 1 = 0 \]
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 8: Determine valid solutions
This gives us two potential solutions for \( y \):
\[ y_1 = \frac{-1 + \sqrt{5}}{2} \quad \text{and} \quad y_2 = \frac{-1 - \sqrt{5}}{2} \]
Since \( y = x^2 \) must be non-negative, we discard \( y_2 \) because it is negative. Thus, we have:
\[ x^2 = \frac{-1 + \sqrt{5}}{2} \]

### Final Answer
The value of \( x^2 \) is:
\[ \boxed{\frac{-1 + \sqrt{5}}{2}} \]