A) \[a\cos \frac{\alpha -\beta }{2}\]
B) \[2a\cos \frac{\alpha -\beta }{2}\]
C) \[a\sin \frac{\alpha -\beta }{2}\]
D) \[2a\sin \frac{\alpha -\beta }{2}\]
Correct Answer: D
Solution :
Distance \[=\sqrt{{{a}^{2}}{{(\cos \alpha -\cos \beta )}^{2}}+{{a}^{2}}{{(\sin \alpha -\sin \beta )}^{2}}}\] \[=a\sqrt{{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\sin }^{2}}\beta -2\cos \alpha \cos \beta -2\sin \alpha \sin \beta }\] \[=a\sqrt{2\left\{ 1-\cos \,(\alpha -\beta ) \right\}}=2a\,\sin \,\left( \frac{\alpha -\beta }{2} \right)\] Trick: Put \[a=1,\,\,\alpha =\frac{\pi }{2},\,\beta =\frac{\pi }{6},\] then the points will be (0, 1) and \[\left( \frac{\sqrt{3}}{2},\,\,\frac{1}{2} \right)\]. Obviously, the distance between these two points is 1 which is given by (d). \[\left\{ \because \,\,2a\,\sin \frac{\alpha -\beta }{2}=2\times 1\times \sin \frac{(\pi /2)-(\pi /6)}{2}=2\times \frac{1}{2}=1 \right\}\]
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