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Q.

The area of a triangle with vertices (acosθ1,bsinθ1) and (acosθ2,bsinθ2) and(acosθ3,bsinθ3),(where  θ1≠θ2≠θ3) is equal to

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a

2absinθ1−θ22sinθ2−θ32sinθ3−θ12

b

absinθ1−θ22sinθ2−θ33sinθ3−θ11

c

4absinθ1−θ21sinθ2−θ32sinθ3−θ13

d

6absinθ1−θ21sinθ2−θ32sinθ3−θ13

answer is A.

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Detailed Solution

Δ=12|acosθ1bsinθ11acosθ2bsinθ21acosθ3bsinθ31|    =12ab|cosθ1−cosθ2sinθ1−sinθ20cosθ2−cosθ3sinθ2−sinθ30cosθ3sinθ31|     =2absinθ1−θ22sinθ2−θ32    =2absinθ1−θ22sinθ2−θ32sinθ3−θ12
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