Find the equation of the curve passing through the point $\left( {0,\frac{\pi }{4}} \right)$ whose differential equation is, $\sin x\cos ydx + \cos x\sin ydy = 0$

The differential equation of the given curve is:
$\sin x\cos ydx + \cos x\sin ydy = 0$
$\Rightarrow$ $\frac{{\sin x\cos ydx + \cos x\sin ydy}}{{\cos x\cos y}} = 0$

$\Rightarrow$ $\tan xdx + \tan ydy = 0$
Integrating both sides,

we get:
$\log (\sec x) + \log (\sec y) = \log C$
$\log (\sec x \cdot \sec y) = \log {\rm{C}}$
$\Rightarrow$ $\sec x \cdot \sec y = {\rm{C}}$

….(1)
The curve passes through point $\left( {0,\frac{\pi }{4}} \right)$.
$\therefore$ $1 \times \sqrt 2 = {\rm{C}}$
$\Rightarrow$ ${\rm{C}} = \sqrt 2$

On substituting ${\rm{C}} = \sqrt 2$
$\sec x \cdot \sec y = \sqrt 2$
$\Rightarrow$ $\sec x \cdot \frac{1}{{\cos y}} = \sqrt 2$

$\Rightarrow$ $\cos y = \frac{{\sec x}}{{\sqrt 2 }}$

Hence, the required equation of the curve is $\cos y = \frac{{\sec x}}{{\sqrt 2 }}$.