In a triangle ABC we define `x = tan""(B-C)/2 tan""A/2, y=tan""(C-A)/2 tan""B/2 and z=tan""(A-B)/2 tan""C/2` then the value of `x + y +z` (in terms of x, y, z) is

Options

A. `xyz`
B. `-xyz`
C. `2xyz`
D. none

Answer

To solve the problem, we need to find the value of \( x + y + z \) where:

\[ x = \tan\left(\frac{B-C}{2}\right) \tan\left(\frac{A}{2}\right), \quad y = \tan\left(\frac{C-A}{2}\right) \tan\left(\frac{B}{2}\right), \quad z = \tan\left(\frac{A-B}{2}\right) \tan\left(\frac{C}{2}\right) \]

### Step 1: Rewrite the expressions for \( x, y, z \)

We start by rewriting the expressions using the tangent subtraction formula:

\[ x = \frac{\tan\left(\frac{B}{2}\right) - \tan\left(\frac{C}{2}\right)}{1 + \tan\left(\frac{B}{2}\right)\tan\left(\frac{C}{2}\right)} \cdot \tan\left(\frac{A}{2}\right) \]

Similarly for \( y \) and \( z \):

\[ y = \frac{\tan\left(\frac{C}{2}\right) - \tan\left(\frac{A}{2}\right)}{1 + \tan\left(\frac{C}{2}\right)\tan\left(\frac{A}{2}\right)} \cdot \tan\left(\frac{B}{2}\right) \]

\[ z = \frac{\tan\left(\frac{A}{2}\right) - \tan\left(\frac{B}{2}\right)}{1 + \tan\left(\frac{A}{2}\right)\tan\left(\frac{B}{2}\right)} \cdot \tan\left(\frac{C}{2}\right) \]

### Step 2: Apply the Compendo and Divendo Rule

Using the compendo and divendo rule, we can express \( \frac{x-1}{x+1} \) for each of \( x, y, z \):

\[ \frac{x-1}{x+1} = \frac{B-C}{B+C}, \quad \frac{y-1}{y+1} = \frac{C-A}{C+A}, \quad \frac{z-1}{z+1} = \frac{A-B}{A+B} \]

### Step 3: Multiply the equations

Now, we multiply these three equations together:

\[ \left(\frac{x-1}{x+1}\right) \left(\frac{y-1}{y+1}\right) \left(\frac{z-1}{z+1}\right) = \frac{(B-C)(C-A)(A-B)}{(B+C)(C+A)(A+B)} \]

### Step 4: Simplify the right-hand side

The right-hand side simplifies to:

\[ \frac{-2C}{2B} \cdot \frac{-2A}{2C} \cdot \frac{-2B}{2A} = \frac{2C \cdot 2A \cdot 2B}{2B \cdot 2C \cdot 2A} = 1 \]

### Step 5: Rearranging the equation

From the multiplication, we have:

\[ (x-1)(y-1)(z-1) = (x+1)(y+1)(z+1) \]

### Step 6: Solve for \( x + y + z \)

Now, we can expand both sides and simplify to find \( x + y + z \):

\[ xyz - (xy + xz + yz) + (x + y + z) - 1 = xyz + (xy + xz + yz) + (x + y + z) + 1 \]

This leads us to:

\[ 2(x + y + z) = xyz + 2 \]

Thus, we find:

\[ x + y + z = \frac{xyz + 2}{2} \]

### Final Result

The value of \( x + y + z \) in terms of \( x, y, z \) is:

\[ \boxed{\frac{xyz + 2}{2}} \]