If `y= tan x cot 3x, x in R`, then

Options

A. `(1)/(3) lt y lt 1`
B. `y lt (1)/(3)" or "y gt 3`
C. `(1)/(3) le y le 1`
D. `(1)/(3) le y le 3`

Answer

To solve the problem where \( y = \tan x \cdot \cot 3x \) and \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = \tan x \cdot \cot 3x \]
Using the identity \( \cot \theta = \frac{1}{\tan \theta} \), we can rewrite \( \cot 3x \) as:
\[ y = \tan x \cdot \frac{1}{\tan 3x} = \frac{\tan x}{\tan 3x} \]

### Step 2: Use the identity for \( \tan 3x \)
We can use the triple angle formula for tangent:
\[ \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \]
Let \( k = \tan x \). Then we have:
\[ \tan 3x = \frac{3k - k^3}{1 - 3k^2} \]
Substituting this back into our expression for \( y \):
\[ y = \frac{k}{\frac{3k - k^3}{1 - 3k^2}} = \frac{k(1 - 3k^2)}{3k - k^3} \]

### Step 3: Simplify the expression
We simplify \( y \):
\[ y = \frac{k(1 - 3k^2)}{3k - k^3} = \frac{k - 3k^3}{3k - k^3} \]
This can be rewritten as:
\[ y = \frac{k(1 - 3k^2)}{k(3 - k^2)} \quad \text{(for } k \neq 0\text{)} \]
Thus,
\[ y = \frac{1 - 3k^2}{3 - k^2} \]

### Step 4: Analyze the range of \( y \)
To find the range of \( y \), we need to analyze the expression:
\[ y = \frac{1 - 3k^2}{3 - k^2} \]
We will find the values of \( y \) for different values of \( k \).

### Step 5: Find critical points
To find the critical points, we can set the derivative of \( y \) with respect to \( k \) to zero. However, we can also analyze the limits:
- As \( k \to 0 \), \( y \to \frac{1}{3} \).
- As \( k^2 \to 3 \), \( y \to \infty \).
- As \( k^2 \to \infty \), \( y \to -3 \).

### Step 6: Determine the intervals
We can find the intervals where \( y \) is positive or negative:
1. Set \( 1 - 3k^2 \geq 0 \) which gives \( k^2 \leq \frac{1}{3} \) or \( k \in \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \).
2. Set \( 3 - k^2 > 0 \) which gives \( k^2 < 3 \) or \( k \in (-\sqrt{3}, \sqrt{3}) \).

### Step 7: Combine intervals
The combined valid interval for \( k \) is:
\[ k \in \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \cap (-\sqrt{3}, \sqrt{3}) = \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \]

### Step 8: Determine the range of \( y \)
From the analysis, we see that:
- \( y \) can take values from \( -\infty \) to \( \frac{1}{3} \) and from \( 3 \) to \( +\infty \).
Thus, the range of \( y \) is:
\[ y \in (-\infty, \frac{1}{3}] \cup [3, +\infty) \]

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