$tan y d x + tan x d y = 0$

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Solution

$t a n y d x + t a n x d y = 0$
$\frac{d x}{t a n x} = - \frac{d y}{t a n y}$
Integrating both side,
$\int \frac{d x}{t a n x} = - \int \frac{d y}{t a n y}$
$\int c o t x d x = - \int c o t y d y$ ......(1)

Now, using substitution -

$\int c o t x d x = \int \frac{c o s x}{s i n x} d x$
Say $u = s i n x .$
then,
$d u = c o s x d x$

Substitute $d u = c o s x, u = s i n x$ in above expression,

$\int \frac{c o s x}{s i n x} d x = \int \frac{d u}{u}$
$= l n | u | + l n C$
Substitute back $u = s i n x$

$\int c o t x d x = \int \frac{c o s x}{s i n x} d x = l n | s i n x | + l n C$

Similarly,
$\int c o t y d y = \int \frac{c o s y}{s i n y} d y = l n | s i n y | + l n C$
Where $C$ is constant.
Substituting these value in expression (1), we get
$l n | s i n x | + l n | s i n y | = l n C$
$l n | s i n x \times s i n y | = l n C$

$| s i n x \times s i n y | = C$
Hence proved.

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