`cos^(-1)x=tan^(-1)x` then

Options

A. `x^(2)=(sqrt(5)-1)/(2)`
B. `x^(2)=(sqrt(5)+1)/(2)`
C. `sin(cos^(-1)x)=(sqrt(5)-1)/(2)`
D. `tan(cos^(-1)x)=(sqrt(5)-1)/(2)`

Answer

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

### Step 1: Set the equations
Let \( \theta = \cos^{-1}x = \tan^{-1}x \). This gives us two equations:
1. \( x = \cos\theta \)
2. \( x = \tan\theta \)

### Step 2: Use the identity relating tan and cos
We know the identity:
\[ \tan\theta = \frac{\sin\theta}{\cos\theta} \]
Using the Pythagorean identity, we have:
\[ \tan^2\theta + 1 = \sec^2\theta \]
This means:
\[ \tan^2\theta = \sec^2\theta - 1 \]

### Step 3: Substitute for \( x \)
From the equations we set, we can express \( \tan\theta \) in terms of \( x \):
\[ \tan^2\theta = x^2 \]
Substituting this into the identity gives:
\[ x^2 = \sec^2\theta - 1 \]

### Step 4: Express secant in terms of cosine
We know that:
\[ \sec\theta = \frac{1}{\cos\theta} \]
Thus:
\[ \sec^2\theta = \frac{1}{\cos^2\theta} \]
Substituting this into the equation gives:
\[ x^2 = \frac{1}{\cos^2\theta} - 1 \]
Rearranging gives:
\[ x^2 + 1 = \frac{1}{\cos^2\theta} \]

### Step 5: Solve for \( \cos^2\theta \)
Taking the reciprocal:
\[ \cos^2\theta = \frac{1}{x^2 + 1} \]

### Step 6: Equate the two expressions for \( \cos\theta \)
From the first equation, we have:
\[ x = \cos\theta \]
Squaring both sides gives:
\[ x^2 = \cos^2\theta \]
Now we can equate the two expressions for \( \cos^2\theta \):
\[ x^2 = \frac{1}{x^2 + 1} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ x^2(x^2 + 1) = 1 \]
Expanding this results in:
\[ x^4 + x^2 - 1 = 0 \]

### Step 8: Let \( y = x^2 \)
Letting \( y = x^2 \), we rewrite the equation as:
\[ y^2 + y - 1 = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 10: Determine valid solutions
Since \( y = x^2 \) must be non-negative, we discard the negative solution:
\[ y = \frac{-1 + \sqrt{5}}{2} \]
Thus:
\[ x^2 = \frac{-1 + \sqrt{5}}{2} \]

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