It is known that sin beta = frac{4}{5} and 0 | vedclass

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(D) Given $\sin \beta = \frac{4}{5}$ and $0 < \beta < \pi$. Since $\cos(\pi/6) = \frac{\sqrt{3}}{2}$,the expression becomes $E = \frac{\sqrt{3} \sin(\

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(D) Given $\sin \beta = \frac{4}{5}$ and $0 Multiplying numerator and denominator by $\sqrt{3}$,we get $E = \frac{3 \sin(\alpha + \beta) - 4 \cos(\alpha + \beta)}{\sqrt{3} \sin \alpha}$.Using $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:If $0 If $\pi/2 < \beta < \pi$,then $\cos \beta = -\frac{3}{5}$. Substituting $\sin \beta = \frac{4}{5}$ and $\cos \beta = -\frac{3}{5}$,we get $E = \frac{3(\sin \alpha \cdot (-\frac{3}{5}) + \cos \alpha \cdot \frac{4}{5}) - 4(\cos \alpha \cdot (-\frac{3}{5}) - \sin \alpha \cdot \frac{4}{5})}{\sqrt{3} \sin \alpha} = \frac{-\frac{9}{5} \sin \alpha + \frac{12}{5} \cos \alpha + \frac{12}{5} \cos \alpha + \frac{16}{5} \sin \alpha}{\sqrt{3} \sin \alpha} = \frac{\frac{7}{5} \sin \alpha + \frac{24}{5} \cos \alpha}{\sqrt{3} \sin \alpha} = \frac{7 + 24 \cot \alpha}{5\sqrt{3}} = \frac{\sqrt{3}(7 + 24 \cot \alpha)}{15}$.

It is known that $\sin \beta = \frac{4}{5}$ and $0 < \beta < \pi$. Then the value of $\frac{\sqrt{3} \sin(\alpha + \beta) - \frac{2}{\cos(\pi/6)} \cos(\alpha + \beta)}{\sin \alpha}$ is:

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