Question

In the interval (0, π/2), area lying between the curves y = tan x and y = cot x and the X-axis is

• A. 4 log 2 sq. units
• B. 3 log 2 sq. units
• C. log 2 sq. units
• D. 2 log 2 sq. units

Hint:

To find the area between two curves, subtract the lower curve from the upper curve and integrate over the given interval. In this case, \( y = \tan x \) is above \( y = \cot x \) in the interval \( (0, \frac{\pi}{2}) \), and the logarithmic results come from the standard integrals of \( \tan x \) and \( \cot x \).

Answer 1

The Correct Option is C

The correct answer is: (C): log 2 sq. units.

We are tasked with finding the area between the curves \( y = \tan x \) and \( y = \cot x \) in the interval \( (0, \frac{\pi}{2}) \), and the X-axis.

Step 1: Set up the integral for the area between the curves

The area between two curves is given by the integral of the difference between the two functions. We need to integrate the difference between \( \tan x \) and \( \cot x \) over the interval \( (0, \frac{\pi}{2}) \). The area is therefore:

\( A = \int_0^{\frac{\pi}{2}} (\tan x - \cot x) \, dx \)

Step 2: Break the integral into two parts

We can separate the integral into two parts: one for \( \tan x \) and one for \( \cot x \):

\( A = \int_0^{\frac{\pi}{2}} \tan x \, dx - \int_0^{\frac{\pi}{2}} \cot x \, dx \)

Step 3: Compute the integrals

We know that the integral of \( \tan x \) is \( -\ln|\cos x| \) and the integral of \( \cot x \) is \( \ln|\sin x| \). So we have:

\( \int \tan x \, dx = -\ln|\cos x| \)

\( \int \cot x \, dx = \ln|\sin x| \)

Step 4: Evaluate the integrals

Now, substitute the limits \( x = 0 \) and \( x = \frac{\pi}{2} \) into both integrals:

\( A = \left[ -\ln|\cos x| \right]_0^{\frac{\pi}{2}} - \left[ \ln|\sin x| \right]_0^{\frac{\pi}{2}} \)

Evaluating each integral at the limits gives:

\( A = \left( -\ln|\cos \frac{\pi}{2}| + \ln|\cos 0| \right) - \left( \ln|\sin \frac{\pi}{2}| - \ln|\sin 0| \right) \)

At \( x = 0 \), \( \cos 0 = 1 \) and \( \sin 0 = 0 \), and at \( x = \frac{\pi}{2} \), \( \cos \frac{\pi}{2} = 0 \) and \( \sin \frac{\pi}{2} = 1 \). Substituting these values gives:

\( A = \left( -\ln(0) + \ln(1) \right) - \left( \ln(1) - \ln(0) \right) \)

Thus, we find that the area is equal to:

\( A = \ln 2 \)

Conclusion:
The area between the curves \( y = \tan x \) and \( y = \cot x \) in the interval \( (0, \frac{\pi}{2}) \) is \( \ln 2 \) square units, so the correct answer is (C): log 2 sq. units.