1. Stating the problem: Prove the trigonometric identities: (a) $\tan \theta \sin \theta + \cos \theta = \sec \theta$ (b) $\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x$ 2. Proving (a): Recall: $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$. Substitute $\tan \theta$: $$\tan \theta \sin \theta + \cos \theta = \frac{\sin \theta}{\cos \theta} \cdot \sin \theta + \cos \theta = \frac{\sin^2 \theta}{\cos \theta} + \cos \theta.$$ Write $\cos \theta$ as $\frac{\cos^2 \theta}{\cos \theta}$ to combine: $$\frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta.$$ Thus, (a) is proven. 3. Proving (b): Recall the double-angle identity: $\sin 2x = 2 \sin x \cos x$, and $\tan x = \frac{\sin x}{\cos x}$. Rewrite numerator: $$2 \tan x = 2 \frac{\sin x}{\cos x} = \frac{2 \sin x}{\cos x}.$$ Rewrite denominator: $$1 + \tan^2 x = 1 + \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}.$$ Divide numerator by denominator: $$\frac{2 \tan x}{1 + \tan^2 x} = \frac{\frac{2 \sin x}{\cos x}}{\frac{1}{\cos^2 x}} = 2 \sin x \cos x.$$ By the double-angle identity, this equals $\sin 2x$. Thus, (b) is proven. **Final answers:** (a) $\tan \theta \sin \theta + \cos \theta = \sec \theta$ (b) $\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x$