i am very much thankful to the one who show quick solution for DE of TanX*TanY=C?
If I undestand correctly You require the DE the solution of which is, in implicit form,...
$\displaystyle \tan x\cdot \tan\ y = c$ (1)
In that case we first separe the variables...
$\displaystyle \tan y= c\cdot \cot x$ (2)
...then compute the logarithm of both sides of (2)...
$\displaystyle \ln \tan y = \ln \cot x + \ln c$ (3)
... then differenziate both sides of (3)...
$\displaystyle \frac{dy}{\tan y\cdot \cos^{2}y}= - \frac{dx}{\cot x\cdot \sin^{2} x}$ (4)
... and finally we obtain...
$\displaystyle y^{'} = -\frac {\sin y\cdot \cos y}{\sin x\cdot cos x}= -\frac{\sin 2y}{\sin 2x}$ (5)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$