In the interval (0, pi / 2),the area lying be | vedclass

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(C) The curves are $y = 	an x$ and $y = \cot x$. They intersect where $	an x = \cot x$,which implies $	an^2 x = 1$,so $	an x = 1$ (since $x \in (0

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$\log 2$ sq units

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(C) The curves are $y = 	an x$ and $y = \cot x$. They intersect where $	an x = \cot x$,which implies $	an^2 x = 1$,so $	an x = 1$ (since $x \in (0, \pi / 2)$),giving $x = \pi / 4$.The area bounded by the curves $y = 	an x$,$y = \cot x$ and the $X$-axis in the interval $(0, \pi / 2)$ is the sum of two parts:$1$. From $x = 0$ to $x = \pi / 4$,the area is bounded by $y = 	an x$ and the $X$-axis.$2$. From $x = \pi / 4$ to $x = \pi / 2$,the area is bounded by $y = \cot x$ and the $X$-axis.Required Area $= \int_0^{\pi / 4} 	an x \, dx + \int_{\pi / 4}^{\pi / 2} \cot x \, dx$$= [\log |\sec x|]_0^{\pi / 4} + [\log |\sin x|]_{\pi / 4}^{\pi / 2}$$= (\log \sec(\pi / 4) - \log \sec 0) + (\log \sin(\pi / 2) - \log \sin(\pi / 4))$$= (\log \sqrt{2} - \log 1) + (\log 1 - \log(1 / \sqrt{2}))$$= \log \sqrt{2} + \log \sqrt{2} = 2 \log \sqrt{2} = 2 \cdot \frac{1}{2} \log 2 = \log 2 	ext{ sq units.}$

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

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