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Question

The sum of the series tan112+tan118+tan1118+tan1132++ is


A


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B


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C


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D

None of these

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Solution

The correct option is B



tan1(12r2)=tan1(24r2)
=tan1(2r+1)(2r1)1+(2r+1)(2r1)
=tan1(2r+1)tan1(2r1)
Thus,
nr=1tan1(12r2)=nr=1tan1(2r+1)tan1(2r1)
=tan1(2n+1)tan1(1)
=tan1(2n+1)π4
limntan1(2n+1)π4
=π2π4=π4


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