The order and degree of the differential equation $\sqrt{\sin \left(x\right)}\left(dx+dy\right)=\sqrt{\cos \left(x\right)}\left(dx-dy\right)$ is

The correct option is C

$\left(1,1\right)$

Explanation for the correct option

The order of a differential equation is defined as the highest order of derivative of the differential equation.

The degree of the differential equation is defined as the power of the highest order of derivative of the differential equation.

Given differential equation is,

$$\begin{gathered} \sqrt{\sin \left(x\right)}\left(dx+dy\right)=\sqrt{\cos \left(x\right)}\left(dx-dy\right) \\ \Rightarrow \left(\sqrt{\sin \left(x\right)}+\sqrt{\cos \left(x\right)}\right)dy=\left(\sqrt{\cos \left(x\right)}-\sqrt{\sin \left(x\right)}\right)dx \\ \Rightarrow \frac{dy}{dx}=\frac{\sqrt{\cos \left(x\right)}-\sqrt{\sin \left(x\right)}}{\sqrt{\sin \left(x\right)}+\sqrt{\cos \left(x\right)}} \end{gathered}$$

Therefore Order$=1$, Degree= $1$

Hence, the correct option is $\left(\mathrm{C}\right)$,