<s>[INST] Solve $\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$ 
How to solve the differential equation 
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y)$$

Hints please.
 [/INST] Let's pose $y = \arcsin(xz)$. Then:
$$\begin{cases}
\displaystyle\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = \frac{x}{\sqrt{1-x^2z^2}}\frac{dz}{dx}\\
\tan(y) = \tan(\arcsin(xz)) = \displaystyle\frac{xz}{\sqrt{1-x^2z^2}}\\
\sin(y) = \sin(\arcsin(xz)) = xz
\end{cases}$$
Then:
$$\frac{dy}{dx} + \frac{1}{x} \tan(y)= \frac{1}{x^2} \tan(y)\sin(y) \Rightarrow\\
\frac{x}{\sqrt{1-x^2z^2}}\frac{dz}{dx} + \frac{1}{x} \frac{xz}{\sqrt{1-x^2z^2}}= \frac{1}{x^2} \frac{xz}{\sqrt{1-x^2z^2}}xz \Rightarrow\\
\frac{dz}{dx} + \frac{1}{x}z = \frac{1}{x}z^2.$$
This is a Bernoulli differential equation with $Q(x) = P(x) = \frac{1}{x}$. The solution is $$z(x) = \frac{1}{ax + 1}.$$
Finally:
$$y(x) = \arcsin\left(\frac{x}{ax + 1}\right)$$
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