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Question

If f(x)=tan1(secx+tanx),π2<x<π2, and f(0)=0, then f(1) is equal to:

A
π+14
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B
π+24
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C
14
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D
π14
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Solution

The correct option is A π+14
f(x)=tan1(secx+tanx)f(x)=tan1(1+sinxcosx)f(x)=tan1⎜ ⎜ ⎜ ⎜ ⎜(cosx2+sinx2)2cos2x2sin2x2⎟ ⎟ ⎟ ⎟ ⎟f(x)=tan1⎢ ⎢1+tanx21tanx2⎥ ⎥f(x)=tan1[tan(π4+x2)]f(x)=π4+x2f(x)=π4x+x24+cf(0)=0c=0f(1)=π4+14=π+14

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