Given that for all real x, the expression `(x^(2)-2x+4)/(x^(2)+2x+4)` lies between `(1)/(3)` and 3 the values between which the expression `(9tan^(2)x+6tanx+4)/(9tan^(2)x-6tanx+4)` lies are

Options

A. 0 and 2
B. -1 and 1
C. -2 and 0
D. `1//3` and 3

Answer

To solve the problem, we need to analyze the given expressions and find the range of the second expression based on the range of the first expression.

### Step-by-Step Solution:

1. **Identify the Given Expression**:
We have the expression:
\[ \frac{x^2 - 2x + 4}{x^2 + 2x + 4} \]
which lies between \(\frac{1}{3}\) and \(3\).

2. **Set the Expression Equal to y**:
Let's denote:
\[ y = \frac{x^2 - 2x + 4}{x^2 + 2x + 4} \]
We need to solve for \(y\) in the range \(\frac{1}{3} < y < 3\).

3. **Cross-Multiply**:
Rearranging gives:
\[ y(x^2 + 2x + 4) = x^2 - 2x + 4 \]
This can be rewritten as:
\[ yx^2 + 2yx + 4y - x^2 + 2x - 4 = 0 \]
Rearranging this gives:
\[ (y - 1)x^2 + (2y + 2)x + (4y - 4) = 0 \]

4. **Apply the Discriminant Condition**:
For \(x\) to be real, the discriminant must be non-negative:
\[ D = (2y + 2)^2 - 4(y - 1)(4y - 4) \geq 0 \]

5. **Calculate the Discriminant**:
Expanding the discriminant:
\[ D = (2y + 2)^2 - 4[(y - 1)(4y - 4)] \]
Simplifying gives:
\[ D = 4y^2 + 8y + 4 - 16y + 16 \]
\[ D = 4y^2 - 8y + 20 \]

6. **Set the Discriminant Greater Than or Equal to Zero**:
We need:
\[ 4y^2 - 8y + 20 \geq 0 \]
The discriminant of this quadratic must also be non-negative:
\[ (-8)^2 - 4 \cdot 4 \cdot 20 \leq 0 \]
This simplifies to:
\[ 64 - 320 < 0 \]
Thus, the quadratic is always positive.

7. **Find the Range for y**:
Since the original expression lies between \(\frac{1}{3}\) and \(3\), we can conclude that:
\[ y \in \left(\frac{1}{3}, 3\right) \]

8. **Substituting for the Second Expression**:
Now we need to analyze the second expression:
\[ \frac{9\tan^2 x + 6\tan x + 4}{9\tan^2 x - 6\tan x + 4} \]
Let \(t = \tan x\). Then we can rewrite it as:
\[ \frac{9t^2 + 6t + 4}{9t^2 - 6t + 4} \]

9. **Setting the New Expression Equal to y**:
Set:
\[ y = \frac{9t^2 + 6t + 4}{9t^2 - 6t + 4} \]
We need to find the range of this expression.

10. **Cross-Multiply Again**:
Rearranging gives:
\[ y(9t^2 - 6t + 4) = 9t^2 + 6t + 4 \]
This leads to:
\[ (9y - 9)t^2 + (-6y - 6)t + (4y - 4) = 0 \]

11. **Apply the Discriminant Condition Again**:
For \(t\) to be real:
\[ D = (-6y - 6)^2 - 4(9y - 9)(4y - 4) \geq 0 \]

12. **Solve the Discriminant**:
After simplifying, we find that the discriminant must also satisfy a similar condition as before.

13. **Conclusion**:
Since the original expression \(y\) lies between \(\frac{1}{3}\) and \(3\), the same will apply to the second expression. Therefore, the values between which the expression \( \frac{9\tan^2 x + 6\tan x + 4}{9\tan^2 x - 6\tan x + 4} \) lies are:
\[ \left(\frac{1}{3}, 3\right) \]

### Final Answer:
The expression lies between \(\frac{1}{3}\) and \(3\).