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Question

If sin1x2+2x+1+sec1x2+2x+1=π2,x0 then the value of 2sec1(x2)+sin1(x2) is equal to

A
3π2
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B
3π2
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C
π2
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D
π2
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Solution

The correct option is B 3π2


Given,sin1x2+2x+1+sec1x2+2x+1=π2

sin1(x+1)2+sec1(x+1)2=π2

sin1(x+1)+sec1(x+1)=π2

sec1(x+1)=π2sin1(x+1)
sec1(x+1)=cos1(x+1)
cos1(1x+1)=cos1(x+1)

1x+1=x+1
(x+1)2=1
x+1=±1
x=0 and x=2
However as cos1(x+1) to be defined it should lie between [1,1]

1(x+1)1
2x0
2sec1(1)+sin1(1) ...since x0
=2ππ2
=3π2

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