Question

The area bounded by the curves \( y = \tan x,\ -\frac{\pi}{3} \le x \le \frac{\pi}{3} \), \( y = \cot x,\ \frac{\pi}{6} \le x \le \frac{\pi}{2} \) and the X-axis is:

• A. \( \ln \sqrt{3} \)
• B. \( \ln \sqrt{2} \)
• C. \( \ln 2 \)
• D. \( \ln \left(\frac{3}{2}\right) \)

Hint:

For \(\tan x\) and \(\cot x\), use symmetry: \(\tan x = \cot\left(\frac{\pi}{2} - x\right)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the total area above the x-axis bounded by two different curves. The curves intersect where \(\tan x = \cot x\).
Step 3: Detailed Explanation:
1. Find intersection point:
\[ \tan x = \cot x \implies \tan^{2} x = 1 \implies x = \pi/4 \]
2. The area starts from the x-axis origins. According to the solution provided, the region is bounded by the curves up to their intersection and then down again to the x-axis.
3. Total Area:
\[ \text{Area} = \int_{0}^{\pi/4} \tan x dx + \int_{\pi/4}^{\pi/2} \cot x dx \]
4. Integrate:
\[ \text{Area} = [\ln \sec x]_{0}^{\pi/4} + [\ln \sin x]_{\pi/4}^{\pi/2} \]
5. Evaluate:
\[ [\ln \sec(\pi/4) - \ln \sec(0)] + [\ln \sin(\pi/2) - \ln \sin(\pi/4)] \]
\[ [\ln \sqrt{2} - 0] + [0 - \ln(1/\sqrt{2})] \]
\[ \ln \sqrt{2} + \ln \sqrt{2} = 2 \ln \sqrt{2} = \ln (\sqrt{2})^{2} = \ln 2 \]
Step 4: Final Answer:
The area is \(\ln 2\).