Let y = y (x) be the solution of the differential equation ^ 2 xdx...

Given,

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

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

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

$⇒ \dfrac{dt}{dy} + {e}^{2y} \times {t}^{2} + t = 0$

$⇒ \dfrac{dt}{dy} + t = - {t}^{2} . {e}^{2y}$

$⇒ \dfrac{1}{{t}^{2}}\dfrac{dt}{dy} + \dfrac{1}{t} = - {e}^{2y}$

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

$⇒ \dfrac{- du}{dy} + u = - {e}^{2y}$

$⇒ \dfrac{du}{dy} - u = {e}^{2y}$

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

So, the solution is given by,

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

$⇒ \dfrac{{e}^{- y}}{t} = {e}^{y} + c$

$⇒ \dfrac{1}{\tan x} \times {e}^{- y} = {e}^{y} + c$

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

$⇒ c = 0$

$⇒ \dfrac{1}{\tan x} = {e}^{2y}$

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

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

$⇒ {e}^{8\alpha} = 9$