If `tan1^(@).tan2^(@).tan^(@)"………tan89^(@) = 4x^(2) + 2x` then the value of `x` is

Options

A. `sin"(pi)/(5)`
B. `cos"(pi)/(5)`
C. `sin"(pi)/(10)`
D. `cos"(4pi)/(5)`

Explanation

To solve the equation \( \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ = 4x^2 + 2x \), we will follow these steps:

### Step 1: Simplify the Left-Hand Side (LHS)
We know that:
\[ \tan(90^\circ - \theta) = \cot(\theta) \]
Using this property, we can pair the terms:
\[ \tan 1^\circ \cdot \tan 89^\circ, \tan 2^\circ \cdot \tan 88^\circ, \ldots, \tan 44^\circ \cdot \tan 46^\circ \]
The middle term is \( \tan 45^\circ \).

### Step 2: Calculate the Product
Each pair can be simplified:
\[ \tan \theta \cdot \tan(90^\circ - \theta) = \tan \theta \cdot \cot \theta = 1 \]
Thus, we have:
\[ (\tan 1^\circ \cdot \tan 89^\circ) \cdots (\tan 44^\circ \cdot \tan 46^\circ) \cdot \tan 45^\circ = 1^{44} \cdot 1 = 1 \]
So, the left-hand side simplifies to:
\[ \tan 1^\circ \cdot \tan 2^\circ \cdots \tan 89^\circ = 1 \]

### Step 3: Set the Equation
Now we equate the left-hand side to the right-hand side:
\[ 1 = 4x^2 + 2x \]

### Step 4: Rearrange the Equation
Rearranging gives:
\[ 4x^2 + 2x - 1 = 0 \]

### Step 5: Apply the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 4, b = 2, c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot (-1) = 4 + 16 = 20 \]

Now substituting into the formula:
\[ x = \frac{-2 \pm \sqrt{20}}{2 \cdot 4} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \]

### Step 6: Final Values of \( x \)
Thus, the values of \( x \) are:
\[ x = \frac{-1 + \sqrt{5}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{5}}{4} \]

### Summary of Solutions
The two values of \( x \) can be expressed as:
\[ x_1 = \frac{\sqrt{5} - 1}{4}, \quad x_2 = \frac{-1 - \sqrt{5}}{4} \]