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Question

The sum of the series 11.3+15.7+19.11+...to is

A
π2
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B
π4
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C
1
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D
π8
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Solution

The correct option is D π8
Given series: Θ=tanΘ13tan3Θ+15tan5Θ17tan7Θ+19tan9Θ111tan11Θ+...
Substitute Θ=π4
π4=(113)+(1517)+(19111)+...
21.3+25.7+29.11+...=π4
Therefore, 11.3+15.7+19.11+...=π8
Ans: D

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