Step-by-Step Solution

1. Given differential equation: \(e^{x}\tan y~dx+(1-e^{x})\sec^{2}y~dy=0\)

2. Rearrange the equation to separate variables:

   \(e^{x}\tan y~dx = -(1-e^{x})\sec^{2}y~dy\)

   \(\frac{e^{x}}{1-e^{x}}dx = -\frac{\sec^{2}y}{\tan y}dy\)

3. Integrate both sides:

   \(\int \frac{e^{x}}{1-e^{x}}dx = -\int \frac{\sec^{2}y}{\tan y}dy\)

4. For the left side, let \(u = 1-e^{x}\), then \(du = -e^{x}dx\). So, \(\int \frac{e^{x}}{1-e^{x}}dx = -\int \frac{1}{u}du = -\ln|u| = -\ln|1-e^{x}|\)

5. For the right side, let \(v = \tan y\), then \(dv = \sec^{2}y~dy\). So, \(-\int \frac{\sec^{2}y}{\tan y}dy = -\int \frac{1}{v}dv = -\ln|v| = -\ln|\tan y|\)

6. Combine the results:

   -\(\ln|1-e^{x}| = -\ln|\tan y| + C\)

   \(\ln|1-e^{x}| = \ln|\tan y| - C\)

   \(\ln|1-e^{x}| - \ln|\tan y| = -C\)

   \(\ln\left|\frac{1-e^{x}}{\tan y}\right| = -C\)

7. Exponentiate both sides:

   \(\frac{1-e^{x}}{\tan y} = e^{-C} = K\) (where K is a constant)

8. Therefore, the general solution is:

   \(1-e^{x} = K\tan y\)