If `tan^(-1)y=4tan^(-1)x`, then `1//y` is zero for

Options

A. `x=1 pm sqrt(2)`
B. `x= sqrt(2) pm sqrt(3)`
C. `x = 3 pm 2 sqrt(2)`
D. All values of x

Answer

To solve the equation \( \tan^{-1} y = 4 \tan^{-1} x \), we will derive the relationship between \( y \) and \( x \) step by step.

### Step 1: Rewrite the equation using a substitution
Let \( \tan^{-1} x = \theta \). Then, we can express \( x \) as:
\[ x = \tan \theta \]
Thus, the original equation becomes:
\[ \tan^{-1} y = 4\theta \]

### Step 2: Express \( y \) in terms of \( \theta \)
Using the identity for tangent of a multiple angle, we have:
\[ y = \tan(4\theta) \]

### Step 3: Use the double angle formula for tangent
We can express \( \tan(4\theta) \) using the double angle formula:
\[ \tan(4\theta) = \frac{2\tan(2\theta)}{1 - \tan^2(2\theta)} \]
Now, we need to find \( \tan(2\theta) \):
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} = \frac{2x}{1 - x^2} \]

### Step 4: Substitute \( \tan(2\theta) \) back into the equation for \( y \)
Now substitute \( \tan(2\theta) \) into the equation for \( y \):
\[ y = \tan(4\theta) = \frac{2 \cdot \frac{2x}{1 - x^2}}{1 - \left(\frac{2x}{1 - x^2}\right)^2} \]

### Step 5: Simplify the expression for \( y \)
First, simplify the denominator:
\[ 1 - \left(\frac{2x}{1 - x^2}\right)^2 = 1 - \frac{4x^2}{(1 - x^2)^2} = \frac{(1 - x^2)^2 - 4x^2}{(1 - x^2)^2} \]
This simplifies to:
\[ (1 - x^2)^2 - 4x^2 = 1 - 2x^2 + x^4 - 4x^2 = 1 - 6x^2 + x^4 \]
Thus, we have:
\[ y = \frac{4x(1 - x^2)}{1 - 6x^2 + x^4} \]

### Step 6: Find \( \frac{1}{y} \)
To find \( \frac{1}{y} \):
\[ \frac{1}{y} = \frac{1 - 6x^2 + x^4}{4x(1 - x^2)} \]
We want to find when \( \frac{1}{y} = 0 \), which occurs when the numerator equals zero:
\[ 1 - 6x^2 + x^4 = 0 \]

### Step 7: Solve the quadratic equation
Let \( t = x^2 \), then we have:
\[ t^2 - 6t + 1 = 0 \]
Using the quadratic formula:
\[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]

### Step 8: Find \( x \)
Since \( t = x^2 \):
\[ x^2 = 3 + 2\sqrt{2} \quad \text{or} \quad x^2 = 3 - 2\sqrt{2} \]
Thus,
\[ x = \pm \sqrt{3 + 2\sqrt{2}} \quad \text{or} \quad x = \pm \sqrt{3 - 2\sqrt{2}} \]

### Conclusion
The values of \( x \) for which \( \frac{1}{y} = 0 \) are:
\[ x = \sqrt{3 + 2\sqrt{2}}, \quad x = -\sqrt{3 + 2\sqrt{2}}, \quad x = \sqrt{3 - 2\sqrt{2}}, \quad x = -\sqrt{3 - 2\sqrt{2}} \]