1. Stating the problem: Solve the differential equation \(\cos(x) \cos(y) \, dx + \sin(x) \sin(y) \, dy = 0\). 2. Rewrite the equation as: $$\cos(x) \cos(y) + \sin(x) \sin(y) \frac{dy}{dx} = 0$$ or $$\frac{dy}{dx} = - \frac{\cos(x) \cos(y)}{\sin(x) \sin(y)} = - \frac{\cos(x)}{\sin(x)} \cdot \frac{\cos(y)}{\sin(y)} = - \cot(x) \cot(y)$$ 3. This can be rearranged as: $$\frac{dy}{dx} = - \cot(x) \cot(y) \implies \frac{dy}{\cot(y)} = - \cot(x) dx$$ 4. Express \(\frac{dy}{\cot(y)}\) in terms of \(y\): Recall \(\cot(y) = \frac{\cos(y)}{\sin(y)}\), so: $$\frac{dy}{\cot(y)} = \frac{dy}{\frac{\cos(y)}{\sin(y)}} = \frac{\sin(y)}{\cos(y)} dy = \tan(y) dy$$ 5. Substitute into the differential form: $$\tan(y) dy = - \cot(x) dx$$ 6. Integrate both sides: $$\int \tan(y) dy = - \int \cot(x) dx$$ 7. Integrate: $$\int \tan(y) dy = - \ln|\cos(y)| + C_1$$ and $$\int \cot(x) dx = \ln|\sin(x)| + C_2$$ 8. So we have: $$- \ln|\cos(y)| = - \ln|\sin(x)| + C$$ or $$\ln|\sin(x)| - \ln|\cos(y)| = C$$ 9. Combine logarithms: $$\ln \left| \frac{\sin(x)}{\cos(y)} \right| = C$$ 10. Exponentiate both sides: $$\left| \frac{\sin(x)}{\cos(y)} \right| = e^{C} = K$$ where \(K\) is an arbitrary positive constant. 11. Therefore, the general solution is: $$\frac{\sin(x)}{\cos(y)} = K$$ or equivalently $$\sin(x) = K \cos(y)$$ --- Problem count: 1