1. **State the problem:** Simplify the expression $$\csc^2 x \tan x - \sin x \sec x.$$\n\n2. **Recall definitions and identities:**\n- $$\csc x = \frac{1}{\sin x}$$\n- $$\sec x = \frac{1}{\cos x}$$\n- $$\tan x = \frac{\sin x}{\cos x}$$\n- Pythagorean identity: $$\csc^2 x = 1 + \cot^2 x$$\n\n3. **Rewrite the expression using definitions:**\n$$\csc^2 x \tan x - \sin x \sec x = \csc^2 x \tan x - \sin x \cdot \frac{1}{\cos x} = \csc^2 x \tan x - \frac{\sin x}{\cos x}.$$\n\n4. **Express $$\csc^2 x$$ and $$\tan x$$ in terms of sine and cosine:**\n$$\csc^2 x \tan x = \left(\frac{1}{\sin^2 x}\right) \left(\frac{\sin x}{\cos x}\right) = \frac{1}{\sin^2 x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x}.$$\n\n5. **Rewrite the entire expression:**\n$$\frac{1}{\sin x \cos x} - \frac{\sin x}{\cos x}.$$\n\n6. **Find common denominator $$\sin x \cos x$$:**\n$$\frac{1}{\sin x \cos x} - \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x} - \frac{\sin x \cdot \sin x}{\cos x \cdot \sin x} = \frac{1}{\sin x \cos x} - \frac{\sin^2 x}{\sin x \cos x}.$$\n\n7. **Combine the fractions:**\n$$\frac{1 - \sin^2 x}{\sin x \cos x}.$$\n\n8. **Use Pythagorean identity $$1 - \sin^2 x = \cos^2 x$$:**\n$$\frac{\cos^2 x}{\sin x \cos x}.$$\n\n9. **Simplify numerator and denominator by canceling $$\cos x$$:**\n$$\frac{\cancel{\cos^2 x}}{\sin x \cancel{\cos x}} = \frac{\cos x}{\sin x} = \cot x.$$\n\n**Final answer:** $$\cot x.$$