1. **State the problem:** Prove that $\tan^2(x) - \sin^2(x) = \sin^4(x) \sec^2(x)$.\n\n2. **Recall definitions and identities:** - $\tan(x) = \frac{\sin(x)}{\cos(x)}$ - $\sec(x) = \frac{1}{\cos(x)}$ - Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$ \n3. **Rewrite the left side using $\tan(x)$:** $$\tan^2(x) - \sin^2(x) = \left(\frac{\sin(x)}{\cos(x)}\right)^2 - \sin^2(x) = \frac{\sin^2(x)}{\cos^2(x)} - \sin^2(x)$$ \n4. **Find common denominator and combine:** $$\frac{\sin^2(x)}{\cos^2(x)} - \sin^2(x) = \frac{\sin^2(x)}{\cos^2(x)} - \frac{\sin^2(x) \cos^2(x)}{\cos^2(x)} = \frac{\sin^2(x) - \sin^2(x) \cos^2(x)}{\cos^2(x)}$$ \n5. **Factor out $\sin^2(x)$ in numerator:** $$\frac{\sin^2(x)(1 - \cos^2(x))}{\cos^2(x)}$$ \n6. **Use Pythagorean identity $1 - \cos^2(x) = \sin^2(x)$:** $$\frac{\sin^2(x) \sin^2(x)}{\cos^2(x)} = \frac{\sin^4(x)}{\cos^2(x)}$$ \n7. **Rewrite denominator using $\sec^2(x) = \frac{1}{\cos^2(x)}$:** $$\sin^4(x) \sec^2(x)$$ \n8. **Conclusion:** Left side simplifies exactly to right side, so $$\tan^2(x) - \sin^2(x) = \sin^4(x) \sec^2(x)$$ which proves the equality.