What is the value of `tan(tan^(-1)x + tan^(-1)y + tan^(-1)z)-cot(cot^(-1)x + cot^(-1)y + cot^(-1)z)` ?

Options

A. 0
B. `2(x + y + z)`
C. `(3 pi)/(2)`
D. `(3 pi)/(2)+ x + y + z`

Explanation

To solve the expression \( \tan(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z) - \cot(\cot^{-1}x + \cot^{-1}y + \cot^{-1}z) \), we can follow these steps:

### Step 1: Simplify the first term
We know that:
\[ \tan(\tan^{-1}a + \tan^{-1}b) = \frac{a + b}{1 - ab} \quad \text{(for } ab < 1\text{)} \]
Using this property, we can find:
\[ \tan(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z) = \tan((\tan^{-1}x + \tan^{-1}y) + \tan^{-1}z) \]
Let \( A = \tan^{-1}x + \tan^{-1}y \), then:
\[ \tan A = \frac{x + y}{1 - xy} \]
Now, we can write:
\[ \tan(A + \tan^{-1}z) = \frac{\tan A + z}{1 - \tan A \cdot z} \]
Substituting for \( \tan A \):
\[ \tan(A + \tan^{-1}z) = \frac{\frac{x + y}{1 - xy} + z}{1 - \frac{x + y}{1 - xy} \cdot z} \]

### Step 2: Simplify the second term
For the second term, we use the identity:
\[ \cot(\cot^{-1}a + \cot^{-1}b) = \frac{ab - 1}{a + b} \quad \text{(for } ab > 1\text{)} \]
Similarly, we can find:
\[ \cot(\cot^{-1}x + \cot^{-1}y + \cot^{-1}z) = \cot((\cot^{-1}x + \cot^{-1}y) + \cot^{-1}z) \]
Let \( B = \cot^{-1}x + \cot^{-1}y \), then:
\[ \cot B = \frac{xy - 1}{x + y} \]
Now, we can write:
\[ \cot(B + \cot^{-1}z) = \frac{\cot B \cdot z - 1}{\cot B + z} \]

### Step 3: Combine the results
Now we have:
\[ \tan(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z) - \cot(\cot^{-1}x + \cot^{-1}y + \cot^{-1}z) \]
Substituting the simplified forms from Steps 1 and 2, we can combine these expressions.

### Final Result
After simplifying the expressions, we find that:
\[ \tan(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z) - \cot(\cot^{-1}x + \cot^{-1}y + \cot^{-1}z) = 0 \]