Question ID: #775
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 $\tan 2\alpha=\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$; $r,s\in\mathbb{N}$ then $r+s$ is equal to
Solution:
We need to find $\tan 2\alpha$. We can express $2\alpha$ as $(\alpha+\beta) + (\alpha-\beta)$.
$$ \tan 2\alpha = \tan[(\alpha+\beta) + (\alpha-\beta)] = \frac{\tan(\alpha+\beta) + \tan(\alpha-\beta)}{1 – \tan(\alpha+\beta)\tan(\alpha-\beta)} $$
**Find $\tan(\alpha+\beta)$:**
Given $\cos(\alpha+\beta) = -\frac{1}{10}$. Since $\alpha \in (0, \frac{\pi}{3})$ and $\beta \in (0, \frac{\pi}{4})$, sum $\alpha+\beta \in (0, \frac{7\pi}{12})$.
Since cosine is negative, $\alpha+\beta$ is in the second quadrant. Tangent will be negative.
$$ \sin(\alpha+\beta) = \sqrt{1 – \left(-\frac{1}{10}\right)^{2}} = \sqrt{\frac{99}{100}} = \frac{3\sqrt{11}}{10} $$
$$ \tan(\alpha+\beta) = \frac{3\sqrt{11}/10}{-1/10} = -3\sqrt{11} $$
**Find $\tan(\alpha-\beta)$:**
Given $\sin(\alpha-\beta) = \frac{3}{8}$.
$$ \cos(\alpha-\beta) = \sqrt{1 – \left(\frac{3}{8}\right)^{2}} = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8} $$
$$ \tan(\alpha-\beta) = \frac{3/8}{\sqrt{55}/8} = \frac{3}{\sqrt{55}} $$
**Calculate $\tan 2\alpha$:**
$$ \tan 2\alpha = \frac{-3\sqrt{11} + \frac{3}{\sqrt{55}}}{1 – (-3\sqrt{11})\left(\frac{3}{\sqrt{55}}\right)} $$
Simplify the numerator:
$$ -3\sqrt{11} + \frac{3}{\sqrt{5}\sqrt{11}} = \frac{-3(11)\sqrt{5} + 3\sqrt{11}}{\sqrt{55}} \text{ … (Rationalizing slightly differently)} $$
Let’s keep the denominator $\sqrt{55}$ common in the numerator term:
$$ \text{Num} = \frac{-3\sqrt{11}\sqrt{55} + 3}{\sqrt{55}} = \frac{-3(11\sqrt{5}) + 3}{\sqrt{55}} = \frac{3(1 – 11\sqrt{5})}{\sqrt{55}} $$
Simplify the denominator of the main fraction:
$$ \text{Denom} = 1 + \frac{9\sqrt{11}}{\sqrt{55}} = 1 + \frac{9\sqrt{11}}{\sqrt{5}\sqrt{11}} = 1 + \frac{9}{\sqrt{5}} = \frac{\sqrt{5} + 9}{\sqrt{5}} $$
Now divide Num by Denom:
$$ \tan 2\alpha = \frac{\frac{3(1 – 11\sqrt{5})}{\sqrt{11}\sqrt{5}}}{\frac{9 + \sqrt{5}}{\sqrt{5}}} = \frac{3(1 – 11\sqrt{5})}{\sqrt{11}(9 + \sqrt{5})} $$
Comparing this with the given form $\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$:
$$ r = 11, \quad s = 9 $$
$$ r + s = 11 + 9 = 20 $$
Ans. (20)
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