If `y=(tanx)/(tan3x),` then

Options

A. `yin[1//3,3]`
B. `y!in[1//3,3]`
C. `yin[-3,-1//3]`
D. `yin[-3,-1//3]`

Answer

To solve the problem where \( y = \frac{\tan x}{\tan 3x} \), we will derive the expression for \( y \) and find its range step by step.

### Step-by-Step Solution:

1. **Write the given expression**:
\[ y = \frac{\tan x}{\tan 3x} \]

2. **Use the formula for \(\tan 3x\)**:
The formula for \(\tan 3x\) is:
\[ \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \]
Substituting this into our expression for \(y\):
\[ y = \frac{\tan x}{\frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}} = \frac{\tan x (1 - 3\tan^2 x)}{3\tan x - \tan^3 x} \]

3. **Simplify the expression**:
Cancel \(\tan x\) from the numerator and denominator (assuming \(\tan x \neq 0\)):
\[ y = \frac{1 - 3\tan^2 x}{3 - \tan^2 x} \]

4. **Let \(t = \tan^2 x\)**:
Substitute \(t\) for \(\tan^2 x\):
\[ y = \frac{1 - 3t}{3 - t} \]

5. **Cross-multiply to eliminate the fraction**:
Rearranging gives:
\[ y(3 - t) = 1 - 3t \]
Expanding this:
\[ 3y - yt = 1 - 3t \]

6. **Rearranging terms**:
Collecting like terms yields:
\[ yt - 3t = 3y - 1 \]
Factoring out \(t\):
\[ t(y - 3) = 3y - 1 \]

7. **Solve for \(t\)**:
Thus, we have:
\[ t = \frac{3y - 1}{y - 3} \]

8. **Since \(t = \tan^2 x\), we know \(t \geq 0\)**:
Therefore, we set up the inequality:
\[ \frac{3y - 1}{y - 3} \geq 0 \]

9. **Determine the critical points**:
The critical points occur when the numerator and denominator are zero:
- \(3y - 1 = 0 \Rightarrow y = \frac{1}{3}\)
- \(y - 3 = 0 \Rightarrow y = 3\)

10. **Test intervals**:
We will test the intervals determined by the critical points:
- For \(y < \frac{1}{3}\): Both \(3y - 1 < 0\) and \(y - 3 < 0\) → Positive
- For \(\frac{1}{3} < y < 3\): \(3y - 1 > 0\) and \(y - 3 < 0\) → Negative
- For \(y > 3\): Both \(3y - 1 > 0\) and \(y - 3 > 0\) → Positive

11. **Conclusion on the range of \(y\)**:
Thus, the solution to the inequality shows that:
\[ y \in (-\infty, \frac{1}{3}) \cup (3, \infty) \]
Therefore, \(y\) does not belong to the intervals \(\left[\frac{1}{3}, 3\right]\).

### Final Answer:
The range of \(y\) is:
\[ y \in (-\infty, \frac{1}{3}) \cup (3, \infty) \]