The solution(s) of the equation `cos^(-1)x=tan^(-1)x` satisfy

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 equation
Let \( y = \tan^{-1} x \). Then we can express \( x \) in terms of \( y \):
\[ x = \tan y \]
Also, from the original equation, we have:
\[ \cos^{-1} x = y \]
This implies:
\[ x = \cos y \]

### Step 2: Equate the two expressions for \( x \)
Since both expressions equal \( x \), we can set them equal to each other:
\[ \tan y = \cos y \]

### Step 3: Rewrite \( \tan y \)
Recall that:
\[ \tan y = \frac{\sin y}{\cos y} \]
Thus, we can rewrite our equation as:
\[ \frac{\sin y}{\cos y} = \cos y \]

### Step 4: Cross-multiply
Cross-multiplying gives us:
\[ \sin y = \cos^2 y \]

### Step 5: Use the Pythagorean identity
Using the identity \( \cos^2 y = 1 - \sin^2 y \), we can substitute:
\[ \sin y = 1 - \sin^2 y \]

### Step 6: Rearrange to form a quadratic equation
Rearranging gives us:
\[ \sin^2 y + \sin y - 1 = 0 \]

### Step 7: Apply the quadratic formula
Using the quadratic formula \( \sin y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):
\[ \sin y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 8: Determine valid solutions for \( \sin y \)
The range of \( \sin y \) is from -1 to 1. Therefore, we need to check the two potential solutions:
1. \( \sin y = \frac{-1 + \sqrt{5}}{2} \)
2. \( \sin y = \frac{-1 - \sqrt{5}}{2} \)

Calculating these:
- For \( \sin y = \frac{-1 + \sqrt{5}}{2} \), this is valid since \( \sqrt{5} \approx 2.236 \), thus \( \frac{-1 + \sqrt{5}}{2} \approx 0.618 \) which is within the range.
- For \( \sin y = \frac{-1 - \sqrt{5}}{2} \), this value is negative and less than -1, hence invalid.

### Step 9: Relate back to \( x \)
Since \( \sin y = x^2 \) (from \( \sin y = \tan^{-1} x \)), we have:
\[ x^2 = \frac{-1 + \sqrt{5}}{2} \]
Thus, the solutions for \( x \) are:
\[ x = \pm \sqrt{\frac{-1 + \sqrt{5}}{2}} \]

### Final Answer
The solutions of the equation \( \cos^{-1} x = \tan^{-1} x \) are:
\[ x = \sqrt{\frac{-1 + \sqrt{5}}{2}} \quad \text{and} \quad x = -\sqrt{\frac{-1 + \sqrt{5}}{2}} \]