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Question

Find the value of the expression 1cos2xsinx+1sin2xcosx, given that 0<x<π2.

A
0
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B
1
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C
2
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D
tanx
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E
sin2x
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Solution

The correct option is B 2
1cos2xsinx+1sin2xcosx=sin2xsinx+cos2xcosx=sinxsinx+cosxcosx=1+1=2

Therefore, Answer is 2

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