1. **State the problem:** Prove the identity $$\frac{\cos x}{\csc^2 x - 1} = \sin x \tan x$$ 2. **Rewrite the denominator:** Recall that $$\csc x = \frac{1}{\sin x}$$ so $$\csc^2 x = \frac{1}{\sin^2 x}$$ Therefore, $$\csc^2 x - 1 = \frac{1}{\sin^2 x} - 1 = \frac{1 - \sin^2 x}{\sin^2 x}$$ 3. **Simplify numerator of denominator:** Using the Pythagorean identity, $$1 - \sin^2 x = \cos^2 x$$ so, $$\csc^2 x - 1 = \frac{\cos^2 x}{\sin^2 x}$$ 4. **Rewrite the left side:** $$\frac{\cos x}{\csc^2 x - 1} = \frac{\cos x}{\frac{\cos^2 x}{\sin^2 x}}$$ 5. **Divide by a fraction:** Dividing by a fraction is multiplying by its reciprocal, $$= \cos x \times \frac{\sin^2 x}{\cos^2 x} = \frac{\cos x \sin^2 x}{\cos^2 x}$$ 6. **Simplify the expression:** Cancel one $\cos x$ from numerator and denominator, $$= \frac{\sin^2 x}{\cos x}$$ 7. **Rewrite the right side:** Note that $$\sin x \tan x = \sin x \times \frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x}$$ 8. **Conclusion:** Both sides equal $$\frac{\sin^2 x}{\cos x}$$ so the identity is proven to be true.