The greatest possible value of the expression `tanx+cotx+cosx` on the interval `[pi/6, pi/4]` is (a) `(12)/(5)sqrt2` (b) `(11)/(6)sqrt2` (c) `(12)/(5)sqrt3` (d) `(11)/(6)sqrt3`

Options

A. `(12)/(5)sqrt2`
B. `(11)/(6)sqrt2`
C. `(12)/(5)sqrt3`
D. `(11)/(6)sqrt3`

Explanation

To find the greatest possible value of the expression \( \tan x + \cot x + \cos x \) on the interval \( \left[\frac{\pi}{6}, \frac{\pi}{4}\right] \), we will follow these steps:

### Step 1: Rewrite the Expression
The expression can be rewritten using trigonometric identities:
\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \]
Thus, the expression becomes:
\[ \tan x + \cot x + \cos x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} + \cos x \]

### Step 2: Combine Terms
Next, we combine the terms under a common denominator:
\[ \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} \]
So, the expression now is:
\[ \frac{1}{\sin x \cos x} + \cos x \]

### Step 3: Use the Identity for \( \sin 2x \)
We know that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can express \( \sin x \cos x \) as:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
This allows us to rewrite the expression as:
\[ \frac{2}{\sin 2x} + \cos x \]

### Step 4: Analyze the Function
Now, we need to analyze the function \( f(x) = \frac{2}{\sin 2x} + \cos x \) over the interval \( \left[\frac{\pi}{6}, \frac{\pi}{4}\right] \).

### Step 5: Determine Monotonicity
To find the maximum value, we need to check the behavior of \( f(x) \) in the given interval. We note that:
- \( \sin 2x \) is increasing in \( \left[\frac{\pi}{6}, \frac{\pi}{4}\right] \).
- \( \cos x \) is decreasing in \( \left[\frac{\pi}{6}, \frac{\pi}{4}\right] \).

Since \( \frac{2}{\sin 2x} \) is increasing and \( \cos x \) is decreasing, the overall function \( f(x) \) is decreasing in this interval.

### Step 6: Evaluate at the Endpoints
To find the maximum value, we evaluate \( f(x) \) at the endpoints of the interval:
1. At \( x = \frac{\pi}{6} \):
\[ f\left(\frac{\pi}{6}\right) = \frac{2}{\sin\left(\frac{\pi}{3}\right)} + \cos\left(\frac{\pi}{6}\right) = \frac{2}{\frac{\sqrt{3}}{2}} + \frac{\sqrt{3}}{2} = \frac{4}{\sqrt{3}} + \frac{\sqrt{3}}{2} \]
Converting to a common denominator:
\[ = \frac{8 + 3}{2\sqrt{3}} = \frac{11}{2\sqrt{3}} \]

2. At \( x = \frac{\pi}{4} \):
\[ f\left(\frac{\pi}{4}\right) = \frac{2}{\sin\left(\frac{\pi}{2}\right)} + \cos\left(\frac{\pi}{4}\right) = 2 + \frac{\sqrt{2}}{2} = 2 + \frac{\sqrt{2}}{2} \]

### Step 7: Compare Values
Now we compare the values obtained:
- \( f\left(\frac{\pi}{6}\right) = \frac{11}{2\sqrt{3}} \)
- \( f\left(\frac{\pi}{4}\right) = 2 + \frac{\sqrt{2}}{2} \)

### Step 8: Conclusion
After evaluating both endpoints, we find that the maximum value occurs at \( x = \frac{\pi}{6} \):
\[ \frac{11}{2\sqrt{3}} = \frac{11\sqrt{3}}{6} \]
Thus, the greatest possible value of the expression \( \tan x + \cot x + \cos x \) on the interval \( \left[\frac{\pi}{6}, \frac{\pi}{4}\right] \) is:
\[ \boxed{\frac{11}{6}\sqrt{3}} \]