1. **Problem a:** Prove that $\sin x \cos x \tan x = 1 - \cos^2 x$. 2. **Recall the identity:** $\tan x = \frac{\sin x}{\cos x}$. 3. Substitute $\tan x$ in the left side: $$\sin x \cos x \tan x = \sin x \cos x \cdot \frac{\sin x}{\cos x}$$ 4. Cancel $\cos x$ in numerator and denominator: $$= \sin x \cancel{\cos x} \cdot \frac{\sin x}{\cancel{\cos x}} = \sin^2 x$$ 5. Use the Pythagorean identity: $$1 - \cos^2 x = \sin^2 x$$ 6. Therefore, $$\sin x \cos x \tan x = \sin^2 x = 1 - \cos^2 x$$ 7. **Problem b:** Prove that $\tan^2 x \sin^2 x = \tan^2 x - \sin^2 x$. 8. Express $\tan^2 x$ as $\frac{\sin^2 x}{\cos^2 x}$: $$\tan^2 x \sin^2 x = \frac{\sin^2 x}{\cos^2 x} \sin^2 x = \frac{\sin^4 x}{\cos^2 x}$$ 9. Rewrite the right side: $$\tan^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x} - \sin^2 x = \frac{\sin^2 x - \sin^2 x \cos^2 x}{\cos^2 x}$$ 10. Factor $\sin^2 x$ in numerator: $$= \frac{\sin^2 x (1 - \cos^2 x)}{\cos^2 x}$$ 11. Use Pythagorean identity $1 - \cos^2 x = \sin^2 x$: $$= \frac{\sin^2 x \sin^2 x}{\cos^2 x} = \frac{\sin^4 x}{\cos^2 x}$$ 12. Both sides equal $\frac{\sin^4 x}{\cos^2 x}$, so the identity holds. **Final answers:** - a) $\sin x \cos x \tan x = 1 - \cos^2 x$ - b) $\tan^2 x \sin^2 x = \tan^2 x - \sin^2 x$