1. **Problem statement:** Prove the identity: $$\frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta} = \frac{2}{\cos \theta}$$ 2. **Start with the left-hand side (LHS):** $$\frac{\cos \theta}{1 + \sin \theta} + \frac{1 + \sin \theta}{\cos \theta}$$ 3. **Find a common denominator:** $$\frac{\cos^2 \theta}{\cos \theta (1 + \sin \theta)} + \frac{(1 + \sin \theta)^2}{\cos \theta (1 + \sin \theta)} = \frac{\cos^2 \theta + (1 + \sin \theta)^2}{\cos \theta (1 + \sin \theta)}$$ 4. **Expand the numerator:** $$(1 + \sin \theta)^2 = 1 + 2 \sin \theta + \sin^2 \theta$$ So numerator: $$\cos^2 \theta + 1 + 2 \sin \theta + \sin^2 \theta$$ 5. **Use the Pythagorean identity:** $$\cos^2 \theta + \sin^2 \theta = 1$$ So numerator becomes: $$1 + 1 + 2 \sin \theta = 2 + 2 \sin \theta = 2(1 + \sin \theta)$$ 6. **Substitute back:** $$\frac{2(1 + \sin \theta)}{\cos \theta (1 + \sin \theta)}$$ 7. **Cancel common factor $1 + \sin \theta$:** $$\frac{2\cancel{(1 + \sin \theta)}}{\cos \theta \cancel{(1 + \sin \theta)}} = \frac{2}{\cos \theta}$$ 8. **This equals the right-hand side (RHS), so the identity is proven.** --- 1. **Problem statement:** Show that: $$\frac{\sin 40^\circ \cdot \cos 70^\circ}{\sin 20^\circ \cdot \cos 50^\circ} + \frac{\sin 10^\circ \cdot \cos 55^\circ}{\sin 35^\circ \cdot \cos 80^\circ} + \frac{\tan 15^\circ \cdot \cot 65^\circ}{\tan 25^\circ \cdot \cot 75^\circ} = 3$$ 2. **Use complementary angle identities:** Note that $\cos 70^\circ = \sin 20^\circ$, $\cos 50^\circ = \sin 40^\circ$, $\cos 55^\circ = \sin 35^\circ$, $\cos 80^\circ = \sin 10^\circ$, $\cot \alpha = \tan (90^\circ - \alpha)$. 3. **Simplify each fraction:** - First fraction: $$\frac{\sin 40^\circ \cdot \cos 70^\circ}{\sin 20^\circ \cdot \cos 50^\circ} = \frac{\sin 40^\circ \cdot \sin 20^\circ}{\sin 20^\circ \cdot \sin 40^\circ} = 1$$ - Second fraction: $$\frac{\sin 10^\circ \cdot \cos 55^\circ}{\sin 35^\circ \cdot \cos 80^\circ} = \frac{\sin 10^\circ \cdot \sin 35^\circ}{\sin 35^\circ \cdot \sin 10^\circ} = 1$$ - Third fraction: $$\frac{\tan 15^\circ \cdot \cot 65^\circ}{\tan 25^\circ \cdot \cot 75^\circ} = \frac{\tan 15^\circ \cdot \tan 25^\circ}{\tan 25^\circ \cdot \tan 15^\circ} = 1$$ 4. **Sum all fractions:** $$1 + 1 + 1 = 3$$ 5. **Hence, the equality is shown.**