If `tan^-1 y=4 tan^-1x, then y` is infinite if

Options

A. `x = 1+ sqrt(3)`
B. `x = 1+ sqrt(2)`
C. `x = 1- sqrt(3)`
D. None of these

Answer

To solve the problem where \( \tan^{-1} y = 4 \tan^{-1} x \) and determine when \( y \) is infinite, we can follow these steps:

### Step 1: Set Up the Equation
Given:
\[ \tan^{-1} y = 4 \tan^{-1} x \]
We know that \( y \) is infinite when \( \tan^{-1} y \) approaches \( \frac{\pi}{2} \).

### Step 2: Substitute the Infinite Value
If \( y \) is infinite, then:
\[ \tan^{-1} y = \frac{\pi}{2} \]
Substituting this into the equation gives:
\[ \frac{\pi}{2} = 4 \tan^{-1} x \]

### Step 3: Solve for \( \tan^{-1} x \)
To isolate \( \tan^{-1} x \), divide both sides by 4:
\[ \tan^{-1} x = \frac{\pi}{8} \]

### Step 4: Find \( x \)
To find \( x \), we take the tangent of both sides:
\[ x = \tan\left(\frac{\pi}{8}\right) \]

### Step 5: Use the Double Angle Formula
Using the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \( \theta = \frac{\pi}{8} \), then \( 2\theta = \frac{\pi}{4} \) and we know \( \tan\left(\frac{\pi}{4}\right) = 1 \):
\[ 1 = \frac{2\tan\left(\frac{\pi}{8}\right)}{1 - \tan^2\left(\frac{\pi}{8}\right)} \]

### Step 6: Set Up the Equation
Let \( t = \tan\left(\frac{\pi}{8}\right) \):
\[ 1 = \frac{2t}{1 - t^2} \]
Cross-multiplying gives:
\[ 1 - t^2 = 2t \]
Rearranging leads to:
\[ t^2 + 2t - 1 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} \]
This simplifies to:
\[ t = -1 \pm \sqrt{2} \]

### Step 8: Choose the Positive Value
Since \( \tan\left(\frac{\pi}{8}\right) \) must be positive, we take:
\[ \tan\left(\frac{\pi}{8}\right) = -1 + \sqrt{2} \]

### Conclusion
Thus, the value of \( x \) is:
\[ x = \sqrt{2} - 1 \]