Let cos(alpha+beta)=-frac{1}{10} and sin(alph | vedclass

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(C) Given $\cos(\alpha+\beta) = -\frac{1}{10}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$, we have $0 < \alpha

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$20$

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(C) Given $\cos(\alpha+\beta) = -\frac{1}{10}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$, we have $0 < \alpha+\beta < \frac{7\pi}{12}$. Since $\cos(\alpha+\beta) < 0$, we must have $\frac{\pi}{2} < \alpha+\beta < \frac{7\pi}{12}$.
Thus, $\sin(\alpha+\beta) = \sqrt{1 - (-\frac{1}{10})^2} = \sqrt{\frac{99}{100}} = \frac{3\sqrt{11}}{10}$.
So, $	an(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11}$.
Given $\sin(\alpha-\beta) = \frac{3}{8}$. Since $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$, we have $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{3}$. Since $\sin(\alpha-\beta) > 0$, we have $0 < \alpha-\beta < \frac{\pi}{3}$.
Thus, $\cos(\alpha-\beta) = \sqrt{1 - (\frac{3}{8})^2} = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8}$.
So, $	an(\alpha-\beta) = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}}$.
Now, $	an 2\alpha = 	an((\alpha+\beta)+(\alpha-\beta)) = \frac{	an(\alpha+\beta) + 	an(\alpha-\beta)}{1 - 	an(\alpha+\beta)	an(\alpha-\beta)}$.
Substituting the values, $	an 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{55}}}{1 - (-3\sqrt{11})(\frac{3}{\sqrt{55}})} = \frac{\frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}}}{1 + \frac{9\sqrt{11}}{\sqrt{55}}} = \frac{-3(11\sqrt{5}) + 3}{\sqrt{55} + 9\sqrt{11}} = \frac{3(1 - 11\sqrt{5})}{\sqrt{11}(\sqrt{5} + 9)}$.
Comparing with $\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$, we get $r=11$ and $s=9$.
Therefore, $r+s = 11+9 = 20$.

Let $\cos(\alpha+\beta)=-\frac{1}{10}$ and $\sin(\alpha-\beta)=\frac{3}{8}$ where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha=\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$, where $r, s \in N$, then $r+s$ is equal to . . . . . .

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