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 \( \tan(x+y) \tan(x-y) = 1 \), we will use the properties of tangent and some trigonometric identities.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \tan(x+y) \tan(x-y) = 1 \]

2. **Use the identity for tangent of sum and difference:**
Recall the identities:
\[ \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
\[ \tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \]

3. **Substitute these identities into the equation:**
Let \( \tan x = a \) and \( \tan y = b \). Then:
\[ \tan(x+y) = \frac{a + b}{1 - ab} \]
\[ \tan(x-y) = \frac{a - b}{1 + ab} \]
Substituting these into the original equation gives:
\[ \left(\frac{a + b}{1 - ab}\right) \left(\frac{a - b}{1 + ab}\right) = 1 \]

4. **Multiply the fractions:**
\[ \frac{(a + b)(a - b)}{(1 - ab)(1 + ab)} = 1 \]
This simplifies to:
\[ \frac{a^2 - b^2}{1 - a^2b^2} = 1 \]

5. **Cross-multiply to eliminate the fraction:**
\[ a^2 - b^2 = 1 - a^2b^2 \]

6. **Rearranging gives:**
\[ a^2 + a^2b^2 - b^2 - 1 = 0 \]

7. **This is a quadratic in \( a^2 \):**
\[ a^2(1 + b^2) - (b^2 + 1) = 0 \]
Factoring gives:
\[ a^2 = \frac{b^2 + 1}{1 + b^2} \]

8. **Since \( \tan(x+y) \tan(x-y) = 1 \), we can also deduce that:**
If \( \tan(x+y) = \cot(x-y) \), then:
\[ x+y = 90^\circ - (x-y) \implies 2x = 90^\circ \implies x = 45^\circ \]

9. **Now, we need to find \( \tan\left(\frac{2x}{3}\right):** \[ \tan\left(\frac{2 \times 45^\circ}{3}\right) = \tan(30^\circ) \] 10. **Using the known value:** \[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \] ### Final Answer: Thus, the value of \( \tan\left(\frac{2x}{3}\right) \) is:
\[ \frac{1}{\sqrt{3}} \]