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Question

The value of tanx1/etdt1+t2+cotx1/edtt(1+t2) is equal to

A
1
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B
1
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C
0
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D
None of these
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Solution

The correct option is B 1
Let I=tanx1/etdt1+t2+cotx1/edtt(1+t2)
On integrating both functions, we get
=12log(1+t2)tanx1/e+{logt12log(1+t2)}cotx1/e
=12{logsec2xlog(1+1e2)}+logcotxlog(1e)12{logcosec2xlog(1+1e2)}
=log(1e)=loge=1

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