Question

(ii) Prove that 1 /pi< 1 To 0 sina pi/1+xsqure dx' <=2\π


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Answer 1

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Answer 2

Correct option is

B

−2secx

Given

2

π

<x<π,

1+sinx

1−sinx

+

1−sinx

1+sinx

=

1+sinx

1−sinx

+

1−sinx

1+sinx

=

(

1+sinx

)(

1−sinx

)

(

1−sinx

)

2

+(

1+sinx

)

2

=

1−sin

2

x

1−sinx+1+sinx

=

cos

2

x

2

=

∣cosx∣

2

As x∈(

2

π

,π),∣cosx∣=−cosx

So

1+sinx

1−sinx

+

1−sinx

1+sinx

=

−cosx

2

=−2secx