1. **State the problem:** Prove the trigonometric identity $$\frac{\cos x}{1-\sin x} = \tan x$$. 2. **Recall the formulas and rules:** - $$\tan x = \frac{\sin x}{\cos x}$$ - Use conjugate multiplication to simplify fractions involving expressions like $$1-\sin x$$. 3. **Start with the left-hand side (LHS):** $$\frac{\cos x}{1-\sin x}$$ Multiply numerator and denominator by the conjugate of the denominator $$1+\sin x$$: $$\frac{\cos x}{1-\sin x} \times \frac{1+\sin x}{1+\sin x} = \frac{\cos x (1+\sin x)}{(1-\sin x)(1+\sin x)}$$ 4. **Simplify the denominator using difference of squares:** $$(1-\sin x)(1+\sin x) = 1 - \sin^2 x$$ 5. **Use the Pythagorean identity:** $$1 - \sin^2 x = \cos^2 x$$ 6. **Substitute back:** $$\frac{\cos x (1+\sin x)}{\cos^2 x}$$ 7. **Simplify the fraction by canceling one $$\cos x$$:** $$\frac{\cancel{\cos x} (1+\sin x)}{\cancel{\cos x} \cos x} = \frac{1+\sin x}{\cos x}$$ 8. **Rewrite the right-hand side (RHS) $$\tan x$$ in terms of sine and cosine:** $$\tan x = \frac{\sin x}{\cos x}$$ 9. **Check if LHS equals RHS:** LHS is $$\frac{1+\sin x}{\cos x}$$, RHS is $$\frac{\sin x}{\cos x}$$. They are not equal as stated, so the original identity $$\frac{\cos x}{1-\sin x} = \tan x$$ is incorrect. 10. **Check if the identity might be $$\frac{\cos x}{1-\sin x} = \sec x + \tan x$$ instead:** Recall that $$\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1+\sin x}{\cos x}$$, which matches our simplified LHS. **Final conclusion:** The correct identity is $$\frac{\cos x}{1-\sin x} = \sec x + \tan x$$, not $$\tan x$$. **Answer:** The given identity is false; the correct identity is $$\frac{\cos x}{1-\sin x} = \sec x + \tan x$$.