1. **State the problem:** Prove the identity $$\frac{\cos\left(\frac{\pi}{2} + x\right)}{\cos(\pi + x)} = \tan x.$$\n\n2. **Recall the trigonometric angle addition formulas:**\n- $$\cos(a + b) = \cos a \cos b - \sin a \sin b$$\n- Also, important special values: $$\cos\left(\frac{\pi}{2} + x\right) = -\sin x$$ and $$\cos(\pi + x) = -\cos x.$$\n\n3. **Evaluate numerator:**\n$$\cos\left(\frac{\pi}{2} + x\right) = -\sin x.$$\n\n4. **Evaluate denominator:**\n$$\cos(\pi + x) = -\cos x.$$\n\n5. **Substitute these into the fraction:**\n$$\frac{\cos\left(\frac{\pi}{2} + x\right)}{\cos(\pi + x)} = \frac{-\sin x}{-\cos x}.$$\n\n6. **Simplify the negatives:**\n$$\frac{\cancel{-}\sin x}{\cancel{-}\cos x} = \frac{\sin x}{\cos x}.$$\n\n7. **Recognize the right side:**\n$$\frac{\sin x}{\cos x} = \tan x.$$\n\n**Final answer:**\n$$\frac{\cos\left(\frac{\pi}{2} + x\right)}{\cos(\pi + x)} = \tan x.$$