Solution of the equation $\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y$ is (where $c$ is arbitrary constant): 

$\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y$

$\Rightarrow \cot y\mathrm{cosec}y\frac{dy}{dx}+\mathrm{cosec}y\cdot \frac{1}{x}=\frac{1}{x^2} ...\left(i\right)$

Put  $\mathrm{cosec}y=v\therefore \mathrm{cosec}y\cot y\frac{dy}{dx}=-\frac{dv}{dx}$

Then, from Eq.(i)  $-\frac{dv}{dx}+\frac{v}{x}=\frac{1}{x^2}$

$\begin{array}{l}\Rightarrow \frac{dv}{dx}-\frac{v}{x}=-\frac{1}{x^2}\\ \therefore \mathrm{IF}=e^{\int -1/xdx}=e^{-\ln x}=e^{\ln (1/x)}=\frac{1}{x}\end{array}$

$\therefore$  Solution is $v\cdot \left(\frac{1}{x}\right)=\int \left(-\frac{1}{x^2}\right)\left(\frac{1}{x}\right)dx+c=\frac{1}{2x^2}+c$

or $\frac{1}{x\sin y}=\frac{1}{2x^2}\left(1+2c{x}^2\right)$

$\Rightarrow 2x=\sin y\left(1+2c{x}^2\right)$

or $2x=\sin y\left(1+c{x}^2\right)(\because c$ is arbitrary constant$)$