Explanation

(a)
Given, $\frac{dy}{dx} - \frac{\tan y}{x} = \frac{\tan y\ \sin y}{x^{2}}$
$\Rightarrow$ $\cot y\ cosec\ y\ \frac{dy}{dx} - \frac{cosec\ y}{x} = \frac{1}{x^{2}}$
Put $- cosec\ y = t$
$\Rightarrow$ $\cot y\ cosec\ y\ \frac{dy}{dx} = \frac{dt}{dx}$
$\therefore$ $\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^{2}}$
$\therefore$ $IF = e^{\int_{}^{}{P\ dx}} = e^{\int_{}^{}{\frac{1}{x}\ dx}} = x$
$\therefore$ Solution is $tx = \int_{}^{}x.\frac{1}{x^{2}}dx - c$
$\Rightarrow$ $- cosec\ y.x = \log x - c$
$\Rightarrow$ $\frac{x}{\sin y} + \log x = c$