If `x+y+z=pi`, `tan x tanz=2` and `tanytanz=18`,then `tan^(2)z=`

Options

A. 15
B. 16
C. 19
D. 20

Explanation

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
- \( x + y + z = \pi \)
- \( \tan x \tan z = 2 \)
- \( \tan y \tan z = 18 \)

2. **Express \( y \) in terms of \( x \) and \( z \):**
\[ y = \pi - x - z \]

3. **Use the tangent addition formula:**
From the equation \( x + y = \pi - z \), we can write:
\[ \tan(x + y) = \tan(\pi - z) = -\tan z \]
Using the tangent addition formula:
\[ \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \]
Therefore, we have:
\[ \frac{\tan x + \tan y}{1 - \tan x \tan y} = -\tan z \]

4. **Substituting known values:**
Substitute \( \tan x \tan z = 2 \) and \( \tan y \tan z = 18 \) into the equation:
\[ \tan y = \frac{18}{\tan z} \]
Substitute this into the tangent addition formula:
\[ \frac{\tan x + \frac{18}{\tan z}}{1 - \tan x \cdot \frac{18}{\tan z}} = -\tan z \]

5. **Cross-multiplying:**
Cross-multiplying gives:
\[ \tan x + \frac{18}{\tan z} = -\tan z \left(1 - \tan x \cdot \frac{18}{\tan z}\right) \]
Simplifying this leads to:
\[ \tan x + \frac{18}{\tan z} = -\tan z + 18 \tan x \]

6. **Rearranging terms:**
Rearranging gives:
\[ \tan x - 18 \tan x + \tan z + \frac{18}{\tan z} = 0 \]
This simplifies to:
\[ -17 \tan x + \tan z + \frac{18}{\tan z} = 0 \]

7. **Multiplying through by \( \tan z \):**
Multiplying through by \( \tan z \) gives:
\[ -17 \tan x \tan z + \tan^2 z + 18 = 0 \]

8. **Substituting \( \tan x \):**
From \( \tan x \tan z = 2 \), we have:
\[ \tan x = \frac{2}{\tan z} \]
Substituting this into the equation:
\[ -17 \left(\frac{2}{\tan z}\right) \tan z + \tan^2 z + 18 = 0 \]
This simplifies to:
\[ -34 + \tan^2 z + 18 = 0 \]
Therefore:
\[ \tan^2 z - 16 = 0 \]

9. **Final result:**
Thus, we find:
\[ \tan^2 z = 16 \]

### Final Answer:
\[ \tan^2 z = 16 \]