1. **Problem 09:** Given $\tan A = 2$ and $\tan B = 3$, find $\frac{\cos(A+B)}{\sin(A-B)}$. 2. Recall the formulas: - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ - $\cos(A+B) = \cos A \cos B - \sin A \sin B$ - $\sin(A-B) = \sin A \cos B - \cos A \sin B$ 3. From $\tan A = 2$, we have $\frac{\sin A}{\cos A} = 2$, so $\sin A = 2 \cos A$. 4. From $\tan B = 3$, we have $\frac{\sin B}{\cos B} = 3$, so $\sin B = 3 \cos B$. 5. Substitute into numerator: $$\cos(A+B) = \cos A \cos B - \sin A \sin B = \cos A \cos B - (2 \cos A)(3 \cos B) = \cos A \cos B - 6 \cos A \cos B = -5 \cos A \cos B$$ 6. Substitute into denominator: $$\sin(A-B) = \sin A \cos B - \cos A \sin B = (2 \cos A) \cos B - \cos A (3 \cos B) = 2 \cos A \cos B - 3 \cos A \cos B = - \cos A \cos B$$ 7. Therefore, $$\frac{\cos(A+B)}{\sin(A-B)} = \frac{-5 \cos A \cos B}{- \cos A \cos B} = 5$$ --- 8. **Problem 10:** Given $\tan \alpha$ and $\tan \beta$ are roots of $x^2 - 4x + 1 = 0$, find $2 \sin^2(\alpha + \beta) + 4 \cos^2(\alpha + \beta)$. 9. From the quadratic, sum and product of roots: $$\tan \alpha + \tan \beta = 4$$ $$\tan \alpha \tan \beta = 1$$ 10. Use the tangent addition formula: $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{4}{1 - 1} = \frac{4}{0}$$ This means $\alpha + \beta = \frac{\pi}{2} + k\pi$, so $\sin(\alpha + \beta) = \pm 1$ and $\cos(\alpha + \beta) = 0$. 11. Calculate: $$2 \sin^2(\alpha + \beta) + 4 \cos^2(\alpha + \beta) = 2 \times 1^2 + 4 \times 0^2 = 2$$ **Final answers:** - Problem 09: $5$ - Problem 10: $2$