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Question

Let $$\alpha,\beta$$ be such that $$0<\alpha-\beta-\pi<2\pi.$$ if $$ \sin \alpha+\sin \beta=-\dfrac{21}{65}$$ and $$\cos \alpha+\cos \beta=-\dfrac{27}{65}.$$ Then the value of cos $$\dfrac{\alpha-\beta}{2}$$ is a


A
665
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B
3130
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C
3130
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D
665
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Solution

The correct option is B $$\dfrac{-3}{\sqrt{130}}$$
$$\left (\sin \alpha +\sin \beta   \right )^{2}+\left (\cos \alpha +\cos \beta   \right )^{2}$$
$$\Rightarrow 2+2 \cos \left (\alpha -\beta   \right )=\dfrac{18}{65}$$
$$\Rightarrow \cos^{2}\dfrac{\alpha -\beta }{2}=\dfrac{9}{130},  As \pi < \alpha -\beta < 3\pi 
\Rightarrow \dfrac{\pi }{2}< \dfrac{\alpha -\beta }{2}< \dfrac{3\pi }{2}$$
$$\Rightarrow \cos \dfrac{\alpha -\beta }{2}=\dfrac{-3}{\sqrt{130}}$$

Mathematics

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