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Question

The value of tanx1/etdt1+t2+cotx1/edtt(1+t2), where x(π6,π3), is

A
0
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B
2
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C
1
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D
3
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Solution

The correct option is C 1
Let f(x)=tanx1/etdt1+t2+cotx1/edtt(1+t2)
f(x)=tanx1+tan2xsec2x+1cotx(1+cot2x)(cosec2x)
f(x)=tanxtanx=0
f(x) is a constant function.
Put x=π4
f(π4)=11etdt(1+t2)+11edtt(1+t2)
f(x)=11e1tdt=[ln|t|]11/e=1

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