If x,y,z are any three real numbers, then ` tan (x-y) + tan (y-z) + tan (z-x)` is equal to

Options

A. 1
B. 0
C. `tan (x-y) tan (y-z) tan (z-x)`
D. `tan (y-x) tan (z-y) tan (z-x)` is equal to

Answer

To solve the problem, we need to find the value of the expression \( \tan(x-y) + \tan(y-z) + \tan(z-x) \).

### Step 1: Use the tangent addition formula
We know that:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
We can apply this to the sum of angles in our expression.

### Step 2: Rewrite the expression
Let:
- \( a = x - y \)
- \( b = y - z \)
- \( c = z - x \)

Thus, we can rewrite the expression as:
\[ \tan(a) + \tan(b) + \tan(c) \]

### Step 3: Recognize the relationship between angles
Notice that:
\[ a + b + c = (x - y) + (y - z) + (z - x) = 0 \]
This means that \( c = -(a + b) \).

### Step 4: Apply the tangent sum formula
Using the identity for tangent of a sum, we have:
\[ \tan(a + b + c) = \tan(0) = 0 \]
Thus,
\[ \tan(a + b) = -\tan(c) \]

### Step 5: Substitute into the tangent identity
Using the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Substituting \( \tan c = -\tan(a + b) \):
\[ 0 = \frac{\tan a + \tan b}{1 - \tan a \tan b} + \tan c \]

### Step 6: Rearranging the equation
This leads to:
\[ \tan a + \tan b + \tan c = \tan a \tan b \tan c \]
Since \( c = -(a + b) \), we can conclude that:
\[ \tan a + \tan b + \tan(-(a + b)) = 0 \]

### Step 7: Conclusion
Thus, we find that:
\[ \tan(x-y) + \tan(y-z) + \tan(z-x) = 0 \]

### Final Answer
The value of \( \tan(x-y) + \tan(y-z) + \tan(z-x) \) is \( 0 \).

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