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

Options

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

Answer

To solve the equation \( \tan(x+y) \tan(x-y) = 1 \), we can use the identity for the tangent of a sum and difference.

### Step 1: Use the tangent addition and subtraction formulas
We know that:
\[ \tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
\[ \tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \]

### Step 2: Substitute into the equation
Substituting these into the original equation gives us:
\[ \left(\frac{\tan x + \tan y}{1 - \tan x \tan y}\right) \left(\frac{\tan x - \tan y}{1 + \tan x \tan y}\right) = 1 \]

### Step 3: Simplify the left-hand side
Multiplying the two fractions, we get:
\[ \frac{(\tan x + \tan y)(\tan x - \tan y)}{(1 - \tan x \tan y)(1 + \tan x \tan y)} = 1 \]
The numerator simplifies to:
\[ \tan^2 x - \tan^2 y \]
And the denominator simplifies to:
\[ 1 - \tan^2 x \tan^2 y \]
So we have:
\[ \frac{\tan^2 x - \tan^2 y}{1 - \tan^2 x \tan^2 y} = 1 \]

### Step 4: Cross-multiply
Cross-multiplying gives us:
\[ \tan^2 x - \tan^2 y = 1 - \tan^2 x \tan^2 y \]

### Step 5: Rearranging the equation
Rearranging this equation, we have:
\[ \tan^2 x + \tan^2 x \tan^2 y = 1 + \tan^2 y \]
Factoring out \( \tan^2 x \) from the left side:
\[ \tan^2 x (1 + \tan^2 y) = 1 + \tan^2 y \]

### Step 6: Solve for \( \tan^2 x \)
Dividing both sides by \( 1 + \tan^2 y \) (assuming \( \tan^2 y \neq -1 \)):
\[ \tan^2 x = \frac{1 + \tan^2 y}{1 + \tan^2 y} = 1 \]

### Step 7: Take the square root
Taking the square root of both sides:
\[ \tan x = 1 \quad \text{or} \quad \tan x = -1 \]

### Conclusion
Since we are looking for the value of \( \tan x \), we conclude:
\[ \tan x = 1 \]