1. **Prove** $\tan \theta \sin \theta + \cos \theta = \sec \theta$. Start from the left side (LHS): LHS = $\tan \theta \sin \theta + \cos \theta$ Recall that $\tan \theta = \frac{\sin \theta}{\cos \theta}$, substitute: LHS = $\frac{\sin \theta}{\cos \theta} \sin \theta + \cos \theta = \frac{\sin^2 \theta}{\cos \theta} + \cos \theta$ Rewrite $\cos \theta$ as $\frac{\cos^2 \theta}{\cos \theta}$ to combine terms: LHS = $\frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta}$ Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$: LHS = $\frac{1}{\cos \theta} = \sec \theta$ This equals the right side (RHS), so the identity holds. 2. **Prove** $\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x$. Start from the left side (LHS): Recall $\tan x = \frac{\sin x}{\cos x}$, substitute: LHS = $\frac{2 \frac{\sin x}{\cos x}}{1 + \left( \frac{\sin x}{\cos x} \right)^2} = \frac{\frac{2 \sin x}{\cos x}}{1 + \frac{\sin^2 x}{\cos^2 x}}$ Simplify denominator: $1 + \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}$ by Pythagorean identity. So LHS equals: $\frac{\frac{2 \sin x}{\cos x}}{\frac{1}{\cos^2 x}} = 2 \sin x \cos x$ Recall the double-angle formula for sine: $\sin 2x = 2 \sin x \cos x$ Thus, LHS = $\sin 2x$, which equals the right side (RHS). Identity proven. **Final answers:** (a) $\tan \theta \sin \theta + \cos \theta = \sec \theta$ (b) $\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x$