1. **Problem 1: Find the minimum value of** $2 \cos \theta + \frac{1}{\sin \theta} + \sqrt{2} \tan \theta$, where $\theta$ is an acute angle. 2. **Using the Arithmetic Mean - Geometric Mean (A.M. ≥ G.M.) inequality:** $$\frac{2 \cos \theta + \frac{1}{\sin \theta} + \sqrt{2} \tan \theta}{3} \geq \sqrt[3]{2 \cos \theta \cdot \frac{1}{\sin \theta} \cdot \sqrt{2} \tan \theta}$$ 3. **Simplify the right side:** $$\sqrt[3]{2 \cos \theta \cdot \frac{1}{\sin \theta} \cdot \sqrt{2} \tan \theta} = \sqrt[3]{2 \cos \theta \cdot \frac{1}{\sin \theta} \cdot \sqrt{2} \cdot \frac{\sin \theta}{\cos \theta}} = \sqrt[3]{2 \cdot \sqrt{2}} = \sqrt[3]{2^{1 + \frac{1}{2}}} = \sqrt[3]{2^{\frac{3}{2}}} = 2^{\frac{1}{2}} = \sqrt{2}$$ 4. **Multiply both sides by 3:** $$2 \cos \theta + \frac{1}{\sin \theta} + \sqrt{2} \tan \theta \geq 3 \sqrt{2}$$ 5. **Conclusion:** The minimum value of the expression is $3 \sqrt{2}$. --- 6. **Problem 2: Prove that for acute angle $\theta$, $\sin \theta < \theta < \tan \theta$.** 7. **Consider a unit circle with center $O$ and radius 1. Let $B$ be the point on the circle at angle 0, and $C$ be the point on the circle at angle $\theta$. Let $D$ be the point on the tangent line at $B$ such that $BD$ is tangent to the circle, and $E$ be the projection of $C$ on $OB$.** 8. **Area of triangle $OBC$ is:** $$\frac{1}{2} \times OB \times CE = \frac{1}{2} \times 1 \times \sin \theta = \frac{\sin \theta}{2}$$ 9. **Area of sector $OBC$ is:** $$\frac{1}{2} \theta$$ 10. **Area of triangle $OBD$ is:** $$\frac{1}{2} \times OB \times BD = \frac{1}{2} \times 1 \times \tan \theta = \frac{\tan \theta}{2}$$ 11. **Since the triangle $OBC$ is inside the sector $OBC$, which is inside triangle $OBD$, we have:** $$\text{Area}(OBC) < \text{Area}(\text{sector } OBC) < \text{Area}(OBD)$$ 12. **Substitute the areas:** $$\frac{\sin \theta}{2} < \frac{\theta}{2} < \frac{\tan \theta}{2}$$ 13. **Multiply all parts by 2:** $$\sin \theta < \theta < \tan \theta$$ 14. **Conclusion:** For acute angle $\theta$, $\sin \theta < \theta < \tan \theta$ is proven.