The area bounded by the curves `y=tanx, -(pi)/(3)le x le(pi)/(3),y=cotx, (pi)/(6)lexle(pi)/(2)` and the x - axis is

Options

A. in `sqrt3`
B. `ln sqrt2`
C. `ln2`
D. `ln((3)/(2))`

Answer

To find the area bounded by the curves \( y = \tan x \) for \( -\frac{\pi}{3} \leq x \leq \frac{\pi}{3} \) and \( y = \cot x \) for \( \frac{\pi}{6} \leq x \leq \frac{\pi}{2} \), we can follow these steps:

### Step 1: Identify the points of intersection
We need to find the points where \( \tan x = \cot x \). This occurs when:
\[ \tan x = \frac{1}{\tan x} \implies \tan^2 x = 1 \implies \tan x = 1 \implies x = \frac{\pi}{4} \]
This point lies within the interval \( \left(\frac{\pi}{6}, \frac{\pi}{2}\right) \).

### Step 2: Determine the area under each curve
The area \( A \) can be calculated as the sum of the areas under the curves from \( \frac{\pi}{6} \) to \( \frac{\pi}{4} \) for \( y = \tan x \) and from \( \frac{\pi}{4} \) to \( \frac{\pi}{3} \) for \( y = \cot x \).

### Step 3: Set up the integrals
The area \( A \) is given by:
\[ A = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \tan x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx \]

### Step 4: Calculate the first integral
The integral of \( \tan x \) is:
\[ \int \tan x \, dx = -\ln |\cos x| + C \]
Thus,
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \tan x \, dx = \left[-\ln |\cos x|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -\ln |\cos(\frac{\pi}{4})| + \ln |\cos(\frac{\pi}{6})| \]
Calculating the values:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
So,
\[ = -\ln\left(\frac{1}{\sqrt{2}}\right) + \ln\left(\frac{\sqrt{3}}{2}\right) = \ln\left(\frac{\sqrt{3}/2}{1/\sqrt{2}}\right) = \ln\left(\frac{\sqrt{3} \cdot \sqrt{2}}{2}\right) = \ln\left(\frac{\sqrt{6}}{2}\right) \]

### Step 5: Calculate the second integral
The integral of \( \cot x \) is:
\[ \int \cot x \, dx = \ln |\sin x| + C \]
Thus,
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot x \, dx = \left[\ln |\sin x|\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \ln |\sin(\frac{\pi}{3})| - \ln |\sin(\frac{\pi}{4})| \]
Calculating the values:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
So,
\[ = \ln\left(\frac{\sqrt{3}/2}{1/\sqrt{2}}\right) = \ln\left(\frac{\sqrt{3} \cdot \sqrt{2}}{2}\right) = \ln\left(\frac{\sqrt{6}}{2}\right) \]

### Step 6: Combine the results
Now we can combine both areas:
\[ A = \ln\left(\frac{\sqrt{6}}{2}\right) + \ln\left(\frac{\sqrt{6}}{2}\right) = 2 \ln\left(\frac{\sqrt{6}}{2}\right) = \ln\left(\left(\frac{\sqrt{6}}{2}\right)^2\right) = \ln\left(\frac{6}{4}\right) = \ln\left(\frac{3}{2}\right) \]

### Final Answer
Thus, the area bounded by the curves and the x-axis is:
\[ \boxed{\ln\left(\frac{3}{2}\right)} \]