The equation of a curve passing through $(1,\frac{\pi}{4})$ and having slope $\frac{\sin 2y}{x+\tan y}$ at (x, y) is:

x = tan y

$\frac{dy}{dx}=\frac{\sin 2y}{x+\tan y}⇒\frac{dx}{dy}=\frac{x}{\sin 2y}+\frac{\tan y}{\sin 2y}⇒\frac{dx}{dy}-\frac{x}{\sin 2y}=\frac{\sec^2y}{2}$

$I.F.=e^{-\int cosec2ydy}=e^{-\frac{1}{2}\log(cosec 2y-\cot 2y)}=e^{-\frac{1}{2}\log(\tan y)}=\frac{1}{\sqrt{\tan y}}$

$x(\frac{1}{\sqrt{\tan y}})=\frac{1}{2}\int\frac{sec^2y}{\sqrt{\tan y}}+dy+c⇒\frac{x}{\sqrt{\tan y}}=\sqrt{\tan y}+c$

As the curve passes through $(1,\frac{\pi}{4})⇒c=0$

$⇒x = tan y$