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Question

If acos3α+3acosαsin2α = m and

asin3α+3acos2αsinα = n, Then (m+n)23+(mn)23

is equal to


A

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B

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C

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D

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Solution

The correct option is C


Adding and subtracting the given relation,

we get (m+ n) = acos3α+3acosαsin2α

+ 3acos2α.sinα+asin3α

= a(cosα+sinα)3

and similarly (m - n) = a(cosαsinα)3

Thus, (m+n)23+(mn)23

= a23 {(cosα+sinα)2 + (cosαsinα)2}

= a23 {2(cos2α+sin2α)} = 2a23.


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