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Question

IfIn=I0π/4tannXdx then 1I2+ I41I3+ I51I5+ I6

A
A . P
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B
G . P
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C
H ., M
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D
None
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Solution

The correct option is B A . P
In=Iπ/40tann1xtan2xdx=Iπ/40tann1xsec2x12xdxIn=[tann+1xn +1]π40In2
In+In2=1n1
Putting n = 4 , 5 , 6 .... we get
I4+I2+I5+I3+I6+I4..... are respectivily 131415 Which are in H . P . are hence their reciprocals are in A . P

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