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Q.

If acos3θ+3acosθsin2θ=m and asin3θ+3acos2θsinθ=n, then (m+n)2/3+(m−n)2/3=

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a

a2/3

b

2a2/3

c

3a2/3

d

4a2/3

answer is B.

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Detailed Solution

m+n=asinθ+cosθ3-3asinθcosθsinθ+cosθ+3asinθcosθ.sinθ+cosθ          =asinθ+cosθ3 similarly,    m-n= asinθ-cosθ3  m+n23+m-n23=2a23
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