What is the solution of the differential equation \[\cos x \cos y \dfrac{{dy}}{{dx}} = - \sin x \sin y \]?
A. \[\sin y + \cos x = c\]
B. \[\sin y - \cos x = c\]
C. \[\sin y \cos x = c\]
D. \[\sin y = c \cos x\]

Hint: Here, the first order differential equation is given. First, simplify the given equation. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve both integrals by using the standard integral rule. In the end, apply the properties of a logarithm to get the solution of the differential equation.

Formula used:
\[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
\[\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x\]
Integration rule: \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \left[ {\log f\left( x \right)} \right] + c\] , where \[c\] is an integration constant.

Complete step by step solution:
The given differential equation is \[\cos x \cos y \dfrac{{dy}}{{dx}} = - \sin x \sin y \].

Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[\dfrac{{\cos y}}{{\sin y}}dy = - \dfrac{{\sin x}}{{\cos x}} dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\dfrac{{\cos y}}{{\sin y}}dy} = \int { - \dfrac{{\sin x}}{{\cos x}} dx} \]
We know the integration rule \[\int {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}dx} = \left[ {\log f\left( x \right)} \right] + c\].
Apply this rule to solve both integrals.
\[\log\left( {\sin y} \right) = \log\left( {\cos x} \right) + \log c\]
Apply the logarithmic property of addition \[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\].
We get,
\[\log\left( {\sin y} \right) = \log\left( {c \cos x} \right)\]
Equating both sides, we get
\[\sin y = c \cos x\]
Hence the correct option is D.

Note: Students often make mistakes when integrating \[ \int {- \dfrac{{\sin x}}{{\cos x}} dx}\]. They take that \[\sin x\] is the derivative of \[\cos x\]. For this reason they get \[\sin y \cos x = c\] as an answer. But the correct answer is \[\sin y = c \cos x\].