Explanation

(b)
Given, $\frac{dy}{dx} = \sin{(x + y)}\tan{(x + y)} - 1$
Put $x + y = z \Rightarrow 1 + \frac{dy}{dx} = \frac{dz}{dx}$
$\therefore$ $\frac{dz}{dx} - 1 = \sin z\tan z - 1$
$\Rightarrow$ $\int_{}^{}\frac{\cos z}{\sin^{2}z}dz = \int_{}^{}{dx}$
Put $\sin z = t$
$\therefore$ $\int_{}^{}\frac{1}{t^{2}}\ dt = x - c \Rightarrow - \frac{1}{t} = x - c$
$\Rightarrow$ $- cosec\ z = x - c$
$\Rightarrow$ $x + cosec\ (x + y) = c$