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Question

The sum of the series upto infinity
tan1(12.12)+tan1(12.22)+tan1(12.32)+... is

A
π4
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B
π2
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C
3π4
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D
π
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Solution

The correct option is A π4
r=1tan1(12r2)=r=1tan1(21+4r21)=r=1tan1((2r+1)(2r1)1+(2r+1)(2r1))=r=1tan1(2r+1)tan1(2r1))
=limn(tan13tan11)+(tan15tan13)+....+(tan1(2n+1)tan1(2n1)
=limn(tan1(2n+1)tan11)=π2π4=π4

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