If π2∫0cotxcotx+cosecxdx=m(π+n), then m.n is equal to :
A
1
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B
−1
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C
12
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D
−12
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Solution
The correct option is B−1 π2∫0cotxcotx+cosecxdx=π2∫0cosxcosx+1dx =π2∫02cos2x2−12cos2x2dx=π2∫0(1−12sec2x2)dx =π2−22tanx2∣∣∣π20 =π2−tanπ4=12⋅(π−2)=m(π+n) ∴m⋅n=12×−2=−1