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Question

Range of f(x)=secx+tanx1tanxsecx+1:xϵ(0,π2) is -

A
(0,1)
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B
(0,)
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C
(1,0)
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D
(,1)
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Solution

The correct option is B (0,)
f(x)=secx+tanx1tanxsecx+1

f(x)=secx+tanx11secx+tanx+1

substitute secx+tanx=u

f(x)=u11u+1

f(x)=u(u1)u+1

u=secx+tanx

u=secxtanx+sec2x>0 in (0,π2)

when x0,u1

when xπ2,u

Therefore the range of f(x)=(0,)

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