If tan (x+y) tan (x-y) = 1, then the value of tan `((2x)/3)` is

Options

A. `1/sqrt(3)`
B. `(2)/(sqrt(3))`
C. `sqrt(3)`
D. `1`

Answer

To solve the equation \( \tan(x+y) \tan(x-y) = 1 \) and find the value of \( \tan\left(\frac{2x}{3}\right) \), we can follow these steps:

### Step 1: Rewrite the equation
Given the equation:
\[ \tan(x+y) \tan(x-y) = 1 \]
This implies:
\[ \tan(x+y) = \cot(x-y) \]
because \( \tan \theta \cdot \cot \theta = 1 \).

### Step 2: Use the cotangent identity
Using the identity \( \cot \theta = \tan(90^\circ - \theta) \), we can rewrite \( \cot(x-y) \):
\[ \tan(x+y) = \tan(90^\circ - (x-y)) \]
This means:
\[ \tan(x+y) = \tan(90^\circ - x + y) \]

### Step 3: Set the angles equal
Since the tangents are equal, we can set the angles equal to each other:
\[ x + y = 90^\circ - x + y \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ x + y = 90^\circ - x + y \]
Subtract \( y \) from both sides:
\[ x = 90^\circ - x \]
Add \( x \) to both sides:
\[ 2x = 90^\circ \]
Divide by 2:
\[ x = 45^\circ \]

### Step 5: Find \( \tan\left(\frac{2x}{3}\right) \)
Now that we have \( x \), we can find \( \tan\left(\frac{2x}{3}\right) \):
\[ \tan\left(\frac{2x}{3}\right) = \tan\left(\frac{2 \cdot 45^\circ}{3}\right) = \tan\left(\frac{90^\circ}{3}\right) = \tan(30^\circ) \]

### Step 6: Calculate \( \tan(30^\circ) \)
We know that:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, the value of \( \tan\left(\frac{2x}{3}\right) \) is:
\[ \frac{1}{\sqrt{3}} \]