1. **State the problem:** Solve the equation $$1 + \frac{\sin x}{\cos x} = \frac{\cos x}{1 - \sin x}$$ for $x$. 2. **Rewrite the equation:** Note that $\frac{\sin x}{\cos x} = \tan x$, so the left side is $1 + \tan x$. 3. **Express both sides with common denominators:** Left side: $$1 + \frac{\sin x}{\cos x} = \frac{\cos x}{\cos x} + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x}$$ Right side is already $$\frac{\cos x}{1 - \sin x}$$ 4. **Set the two fractions equal:** $$\frac{\cos x + \sin x}{\cos x} = \frac{\cos x}{1 - \sin x}$$ 5. **Cross multiply:** $$(\cos x + \sin x)(1 - \sin x) = \cos^2 x$$ 6. **Expand the left side:** $$\cos x (1 - \sin x) + \sin x (1 - \sin x) = \cos^2 x$$ $$\cos x - \cos x \sin x + \sin x - \sin^2 x = \cos^2 x$$ 7. **Group terms:** $$\cos x + \sin x - \cos x \sin x - \sin^2 x = \cos^2 x$$ 8. **Use the Pythagorean identity:** $$\sin^2 x + \cos^2 x = 1 \implies \cos^2 x = 1 - \sin^2 x$$ 9. **Substitute $\cos^2 x$:** $$\cos x + \sin x - \cos x \sin x - \sin^2 x = 1 - \sin^2 x$$ 10. **Add $\sin^2 x$ to both sides:** $$\cos x + \sin x - \cos x \sin x = 1$$ 11. **Rewrite:** $$\cos x + \sin x - \cos x \sin x = 1$$ 12. **Isolate terms:** $$\cos x + \sin x = 1 + \cos x \sin x$$ 13. **Try to find solutions by inspection:** - If $x = 0$, then $\cos 0 = 1$, $\sin 0 = 0$, left side $1 + 0 = 1$, right side $1 + 1*0 = 1$, true. - If $x = \frac{\pi}{2}$, $\cos \frac{\pi}{2} = 0$, $\sin \frac{\pi}{2} = 1$, left side $0 + 1 = 1$, right side $1 + 0*1 = 1$, true. 14. **General solution:** The equation holds for $x = n\pi$ and $x = \frac{\pi}{2} + n\pi$ where $n$ is any integer. **Final answer:** $$x = n\pi \quad \text{or} \quad x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$$