Search any question & find its solution
Question:
Answered & Verified by Expert
In the interval $(0, \pi / 2)$ area lying between the curves $y=\tan x$ and $y=\cot x$ and the $X$-axis is
Options:
Solution:
2007 Upvotes
Verified Answer
The correct answer is:
$\log 2$ sq units
Given, $c_1 \rightarrow y=\tan x, x \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$
$c_2 \rightarrow y=\cot x \in\left[\frac{\pi}{6}, \frac{\pi}{6}\right]$
The region is symmetric about lines $x=\frac{\pi}{4}$ and its first portion is bounded between the lines $x=0$ and $x=\frac{\pi}{4}$
$\therefore$ Required Area $=2 \int_0^{\pi / 4} \tan x d y$
$\begin{aligned} & =2[\log (\sec x)]_0^{\frac{\pi}{4}} \\ & =2\left[\log \left(\sec \frac{\pi}{4}\right)-\log (\sec 0)\right] \\ & =2 \log \sqrt{2}=\log 2 \text { sq units. }\end{aligned}$
$c_2 \rightarrow y=\cot x \in\left[\frac{\pi}{6}, \frac{\pi}{6}\right]$
The region is symmetric about lines $x=\frac{\pi}{4}$ and its first portion is bounded between the lines $x=0$ and $x=\frac{\pi}{4}$
$\therefore$ Required Area $=2 \int_0^{\pi / 4} \tan x d y$
$\begin{aligned} & =2[\log (\sec x)]_0^{\frac{\pi}{4}} \\ & =2\left[\log \left(\sec \frac{\pi}{4}\right)-\log (\sec 0)\right] \\ & =2 \log \sqrt{2}=\log 2 \text { sq units. }\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.