$$\begin{aligned} & \text { If } \mathrm{e}^{\int \tan x d x}=\mathrm{e}^{\ln (\sec x)}=\sec x \\ & \therefore y \cdot \sec x=\int\left\{\frac{2+\sec x}{(1+2 \sec x)^2}\right\} \sec x d x \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =\int \frac{2 \cos x+1}{(\cos x+2)^2} d x \text { Let } \cos x=\frac{1-t^2}{1+t^2} \\ & =\int \frac{2\left(\frac{1-t^2}{1+t^2}\right)+1}{\left(\frac{1-t^2}{1+t^2}+2\right)^2} 2 d t \\ & =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 d t \\ & =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} d t \end{aligned}\\ &\text { Let } \mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}\\ &\left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \end{aligned}$$

$=-2 \int \frac{\mathrm{du}}{\mathrm{u}^2}$

$$\begin{aligned} & y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c\quad\text{..... (I)} \end{aligned}$$

$$\begin{aligned} & \text { At } \mathrm{x}=\frac{\pi}{3}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}} \\ & \text { 2. } \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3 \sqrt{3}}+\mathrm{c} \\ & \text { 2. } \frac{\sqrt{3}}{10}=\frac{2 \sqrt{3}}{10}+\mathrm{c} \Rightarrow \mathrm{C}=0 \\ & \text { At } \mathrm{x}=\frac{\pi}{4}, \mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1 \end{aligned}$$

$$\begin{aligned} & \therefore y \cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}} \\ & y \cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2 \sqrt{2}} \\ & y=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}} \times \frac{2 \sqrt{2}-1}{7} \\ & =\frac{4-\sqrt{2}}{14} \end{aligned}$$