If `xa n dy` are positive acute angles such that `(x+y)` and `(x-y)` satisfy the equation `tan^2theta-4tantheta+1=0,` then `x=pi/6` (b) `y=pi/4` (c) `y=pi/6` (d) `y=pi/4`

Options

A. `x=pi/6`
B. `y=pi/4`
C. `y=pi/6`
D. `x=pi/4`

Explanation

To solve the problem, we need to analyze the given equation and the conditions provided. Let's break down the steps:

### Step 1: Understand the given equation
We are given that \( x+y \) and \( x-y \) satisfy the equation:
\[ \tan^2 \theta - 4 \tan \theta + 1 = 0 \]

### Step 2: Find the roots of the equation
We can use the quadratic formula to find the roots of the equation \( a = 1, b = -4, c = 1 \):
\[ \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ \tan \theta = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \]

### Step 3: Assign the roots to \( \tan(x+y) \) and \( \tan(x-y) \)
Let:
\[ \tan(x+y) = 2 + \sqrt{3} \quad \text{and} \quad \tan(x-y) = 2 - \sqrt{3} \]

### Step 4: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
We can express \( \tan(2x) \) as:
\[ \tan(2x) = \tan(x+y) + \tan(x-y) = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \]
And the product:
\[ \tan(x+y) \tan(x-y) = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \]

### Step 5: Substitute into the tangent formula
Now substituting into the tangent addition formula:
\[ \tan(2x) = \frac{\tan(x+y) + \tan(x-y)}{1 - \tan(x+y) \tan(x-y)} = \frac{4}{1 - 1} = \frac{4}{0} \]
This implies that \( \tan(2x) \) is undefined, which occurs when \( 2x = \frac{\pi}{2} + n\pi \) for some integer \( n \). For the acute angles, we take:
\[ 2x = \frac{\pi}{2} \implies x = \frac{\pi}{4} \]

### Step 6: Find \( y \)
Now we can find \( y \) using the values of \( \tan(x+y) \) and \( \tan(x-y) \):
Using the equation:
\[ \tan(x+y) = 2 + \sqrt{3} \implies x+y = \frac{\pi}{4} + y \]
And since \( \tan(x-y) = 2 - \sqrt{3} \):
\[ x-y = \frac{\pi}{4} - y \]

### Step 7: Solve for \( y \)
Using the equations:
1. \( x + y = \frac{\pi}{4} + y = \tan^{-1}(2 + \sqrt{3}) \)
2. \( x - y = \frac{\pi}{4} - y = \tan^{-1}(2 - \sqrt{3}) \)

We can solve for \( y \):
From the first equation, we can use the known values of \( \tan^{-1}(2 + \sqrt{3}) \) and \( \tan^{-1}(2 - \sqrt{3}) \) to find \( y \).

By solving, we find:
\[ y = \frac{\pi}{6} \]

### Final Answer
Thus, the values are:
- \( x = \frac{\pi}{4} \)
- \( y = \frac{\pi}{6} \)

### Conclusion
The correct option for \( y \) is \( \frac{\pi}{6} \).