Let $y=y(x)$ be the solution of the differential equation $\sec^{2}x \, dx + \left(e^{2y}\tan^{2}x + \tan…

Given,

${\sec }^{2}xdx+\left(e^{2y}{\tan }^{2}x+\tan x\right)dy=0,$

$\Rightarrow {\sec }^2\frac{dx}{dy}+e^{2y}{\tan }^{2}x+\tan x=0$

Now, let $\tan x=t\Rightarrow {\sec }^{2}x\frac{dx}{dy}=\frac{dt}{dy}$

$\Rightarrow \frac{dt}{dy}+e^{2y}\times t^2+t=0$

$\Rightarrow \frac{dt}{dy}+t=-t^2.e^{2y}$

$\Rightarrow \frac{1}{t^2}\frac{dt}{dy}+\frac{1}{t}=-e^{2y}$

Now, taking $\frac{1}{t}=u\Rightarrow \frac{-1}{t^2}\frac{dt}{dy}=\frac{du}{dy}$

$\Rightarrow \frac{-du}{dy}+u=-e^{2y}$

$\Rightarrow \frac{du}{dy}-u=e^{2y}$

Now, finding an integrating factor $\mathrm{IF}=e^{-\int dy}=e^{-y}$

So, the solution is given by,

$u{e}^{-y}=\int e^{-y}\times e^{2y}dy$

$\Rightarrow \frac{e^{-y}}{t}=e^y+c$

$\Rightarrow \frac{1}{\tan x}\times e^{-y}=e^y+c$

Now, using the given condition at $x=\frac{\pi }{4},y=0$ we get,

$\Rightarrow c=0$

$\Rightarrow \frac{1}{\tan x}=e^{2y}$

Now, putting the value $x=\frac{\pi }{6}, y=\alpha$ we get,

$e^{2\alpha }=\sqrt{3}$

$\Rightarrow e^{8\alpha }=9$