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(d) $\sin \alpha + \sin \beta = - \frac{{21}}{{65}},\;\cos \alpha + \cos \beta = - \frac{{27}}{{65}}$
Now ${(\sin \alpha + \sin \beta )^2} + {(\cos \

Vedclass · Accepted Answer

$ - \frac{3}{{\sqrt {130} }}$

Vedclass · Answer

(d) $\sin \alpha + \sin \beta = - \frac{{21}}{{65}},\;\cos \alpha + \cos \beta = - \frac{{27}}{{65}}$
Now ${(\sin \alpha + \sin \beta )^2} + {(\cos \alpha + \cos \beta )^2}$
$ = {\left( {\frac{{ - 21}}{{65}}} \right)^2} + {\left( {\frac{{ - 27}}{{65}}} \right)^2}$
==> $2 + 2\sin \alpha \sin \beta + 2\cos \alpha \cos \beta = \frac{{441}}{{{{65}^2}}} + \frac{{729}}{{{{65}^2}}}$
==> $2 + 2\left[ {\cos (\alpha - \beta )} \right] = \frac{{1170}}{{{{(65)}^2}}} \Rightarrow 2.2{\cos ^2}\left( {\frac{{\alpha + \beta }}{2}} \right) = \frac{{1170}}{{{{(65)}^2}}}$
==> $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = \frac{{3\sqrt {130} }}{{130}} = \frac{3}{{\sqrt {130} }}$
Therefore $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = \frac{{ - 3}}{{\sqrt {130} }}$, .

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