Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=

sin2A

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cosA

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sinA

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cos2A

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Solution

The correct option is B
cosA

sinx siny + cosx cosy = cos(x-y)

Substituting

x=(n+1)A

y=(n+2)A

we get

sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A=cos[(n+1)A-(n+2)A]

=cos[-A]

=cosA

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