1. **State the problem:** Prove the trigonometric identity $$\frac{\cos x}{1-\sin x} - \tan x = \sec x$$. 2. **Recall definitions and formulas:** - $$\tan x = \frac{\sin x}{\cos x}$$ - $$\sec x = \frac{1}{\cos x}$$ - Important rule: To combine terms, get a common denominator. 3. **Rewrite the left side (LHS):** $$\frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x}$$ 4. **Find common denominator:** $$ (1-\sin x)(\cos x) $$ 5. **Express both terms with common denominator:** $$\frac{\cos x \cdot \cos x}{(1-\sin x)\cos x} - \frac{\sin x (1-\sin x)}{(1-\sin x)\cos x}$$ 6. **Simplify numerator:** $$\frac{\cos^2 x - \sin x + \sin^2 x}{(1-\sin x)\cos x}$$ 7. **Use Pythagorean identity:** $$\sin^2 x + \cos^2 x = 1$$ 8. **Replace numerator:** $$\frac{1 - \sin x}{(1-\sin x)\cos x}$$ 9. **Cancel common factor:** $$\frac{\cancel{1 - \sin x}}{\cancel{1-\sin x} \cos x} = \frac{1}{\cos x}$$ 10. **Recognize the result:** $$\frac{1}{\cos x} = \sec x$$ 11. **Conclusion:** LHS equals RHS, so the identity is proven. **Final answer:** $$\frac{\cos x}{1-\sin x} - \tan x = \sec x$$