dy/dx +(1/x) tan y = (1/x²)tany.siny

dy/dx +(1/x)  tan y = (1/x²)tany.siny

divide the eqn. by tan y.sin y

so eqn comes to be

(1/tan y.sin y)*dy/dx + 1/ x.sin y = 1/x²

put 1/sin y = t

(-cosy / sin²y)*dy/dx = dt/dx

on substituting,we get

-dt/dx + t/x = 1/x²
 or dt/dx - t/x = -1/x²

which is of form dy/dx +P(x).y = Q(x)  with P(x) = -1/x and  Q(x) = -1/x²

solution is t.e^{integral P(x)} = integral ( Q(x)*e^{integral P(x)})dx +C

soln is:

t .e ^-lnx =  -1/x^{2 .}e ^-lnx dx

t/x = 1/2x² +C

i.e., 1/x. sin y = 1/(2x²) +C