If x,y,z are variables and `3tan x+4tany+5tanz=20,` then the minimum value of `tan^(2)x+tan^(2)y+tan^(2)z,` is

Options

A. 10
B. 15
C. 8
D. 12

Explanation

To find the minimum value of \( \tan^2 x + \tan^2 y + \tan^2 z \) given the equation \( 3 \tan x + 4 \tan y + 5 \tan z = 20 \), we can use the Cauchy-Schwarz inequality.

### Step-by-Step Solution:

1. **Define the Variables:**
Let \( a = \tan x \), \( b = \tan y \), and \( c = \tan z \). The equation then becomes:
\[ 3a + 4b + 5c = 20 \]

2. **Apply Cauchy-Schwarz Inequality:**
By the Cauchy-Schwarz inequality, we have:
\[ (3a + 4b + 5c)^2 \leq (3^2 + 4^2 + 5^2)(a^2 + b^2 + c^2) \]
Here, \( 3^2 + 4^2 + 5^2 = 9 + 16 + 25 = 50 \).

3. **Substitute the Known Value:**
Substitute \( 3a + 4b + 5c = 20 \) into the inequality:
\[ 20^2 \leq 50(a^2 + b^2 + c^2) \]
This simplifies to:
\[ 400 \leq 50(a^2 + b^2 + c^2) \]

4. **Solve for \( a^2 + b^2 + c^2 \):**
Dividing both sides by 50 gives:
\[ \frac{400}{50} \leq a^2 + b^2 + c^2 \]
Thus:
\[ 8 \leq a^2 + b^2 + c^2 \]

5. **Conclusion:**
Therefore, the minimum value of \( \tan^2 x + \tan^2 y + \tan^2 z \) is:
\[ \tan^2 x + \tan^2 y + \tan^2 z \geq 8 \]
Hence, the minimum value is \( \boxed{8} \).