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(A) The given differential equation is $\frac{d y}{d x}=\frac{	an x+y}{\sin x(\sec x-\sin x 	an x)}$.Simplifying the denominator: $\sin x(\frac{1}{\

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$\sqrt{3}\left(2+\log _{e} \sqrt{3}\right)$

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(A) The given differential equation is $\frac{d y}{d x}=\frac{	an x+y}{\sin x(\sec x-\sin x 	an x)}$.Simplifying the denominator: $\sin x(\frac{1}{\cos x}-\frac{\sin^2 x}{\cos x}) = \sin x(\frac{1-\sin^2 x}{\cos x}) = \sin x(\frac{\cos^2 x}{\cos x}) = \sin x \cos x$.So,$\frac{d y}{d x} = \frac{	an x + y}{\sin x \cos x} = \frac{	an x}{\sin x \cos x} + \frac{y}{\sin x \cos x} = \sec^2 x + y(2 \csc 2x)$.This is a linear differential equation of the form $\frac{d y}{d x} - (2 \csc 2x)y = \sec^2 x$.The integrating factor $I.F. = e^{\int -2 \csc 2x dx} = e^{-\ln|	an x|} = \frac{1}{	an x}$ (since $x \in (0, \pi/2)$).The general solution is $y \cdot \frac{1}{	an x} = \int \sec^2 x \cdot \frac{1}{	an x} dx + c$.Let $	an x = t$,then $\sec^2 x dx = dt$. So,$y \cot x = \int \frac{1}{t} dt + c = \ln|t| + c = \ln(	an x) + c$.Thus,$y = 	an x (\ln(	an x) + c)$.Given $y(\frac{\pi}{4}) = 2$,we have $2 = 	an(\frac{\pi}{4})(\ln(	an(\frac{\pi}{4})) + c) = 1(0 + c)$,so $c = 2$.The solution is $y = 	an x (\ln(	an x) + 2)$.For $x = \frac{\pi}{3}$,$y(\frac{\pi}{3}) = 	an(\frac{\pi}{3})(\ln(	an(\frac{\pi}{3})) + 2) = \sqrt{3}(\ln \sqrt{3} + 2) = \sqrt{3}(\frac{1}{2} \ln 3 + 2) = \sqrt{3}(2 + \log_e \sqrt{3})$.

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}$,$x \in\left(0, \frac{\pi}{2}\right)$ satisfying the condition $y\left(\frac{\pi}{4}\right)=2$. Then,$y\left(\frac{\pi}{3}\right)$ is

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