1. Problem statement: Prove the identity $\tan \theta \sin \theta + \cos \theta = \sec \theta$. 2. Recall definitions: $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$. 3. Substitute $\tan \theta$ into the left side: $$\tan \theta \sin \theta + \cos \theta = \left( \frac{\sin \theta}{\cos \theta} \right) \sin \theta + \cos \theta = \frac{\sin^2 \theta}{\cos \theta} + \cos \theta.$$ 4. Combine terms over common denominator $\cos \theta$: $$\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta}.$$ 5. Use Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$, so $$\frac{1}{\cos \theta} = \sec \theta.$$ 6. Hence, the left side equals the right side: $$\tan \theta \sin \theta + \cos \theta = \sec \theta.$$ --- 7. Problem statement: Prove the identity $\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x$. 8. Recall that $\tan x = \frac{\sin x}{\cos x}$ and $\sin 2x = 2 \sin x \cos x$. 9. Substitute $\tan x$ into the left side: $$\frac{2 \tan x}{1 + \tan^2 x} = \frac{2 \frac{\sin x}{\cos x}}{1 + \left( \frac{\sin x}{\cos x} \right)^2} = \frac{2 \frac{\sin x}{\cos x}}{1 + \frac{\sin^2 x}{\cos^2 x}}.$$ 10. Simplify denominator: $$1 + \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}.$$ 11. Substitute back: $$\frac{2 \frac{\sin x}{\cos x}}{\frac{1}{\cos^2 x}} = 2 \frac{\sin x}{\cos x} \times \cos^2 x = 2 \sin x \cos x.$$ 12. Recognize the right side: $2 \sin x \cos x = \sin 2x$. 13. Hence, $$\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x.$$