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Question

If asin2θ+bsinθ cosθ+c cos2θ12(a+c)12k, then k2 is equal to


Solution

The correct option is A

(a) a sin2 θ+b sin θ cos θ+c cos2 θ12(a+c)
       = 12[a cos 2 θ+b sin 2θ+c cos 2θ]
       = 12[b sin 2 θ(ac)cos 2θ]
       |b sin 2θ(ac)cos 2θ|b2+(ac)2
       12b sin 2θ(ac)cos 2θ12b2+(ac)2
       K=b2+(ac)2

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