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Question

The value of 1+sinA1sinA where A[0,2π]{π2} is 


Your Answer
A
secAtanA if A[0,π2)
Correct Answer
B
(secA+tanA) if A(π2,π]
Your Answer
C
secA+tanA if A[π,3π2)
Correct Answer
D
secA+tanA if A(3π2,2π]

Solution

The correct options are
B (secA+tanA) if A(π2,π]
D secA+tanA if A(3π2,2π]
1+sinA1sinA=(1+sinA)21sin2A                     =|1+sinA||cosA|
                     =1+sinA|cosA|  (1+sinA[0,2])

|cosA|=⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪+cosA,A[0,π2)[3π2,2π]cosA,A(π2,3π2)

If A[0,π2)(3π2,2π] as at A=3π2, cosA=0 
1+sinA1sinA=1+sinAcosA=secA+tanA

If A(π2,3π2)
1+sinA1sinA=1+sinAcosA=(secA+tanA)

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