If 1−sinA1+sinA=secA−tanA then A lies in the quadrants

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I, II

II, III

I, IV

I, III

answer is C.

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Detailed Solution

1−sinA1+sinA=secA−tanA ⇒1−sinA2cos2A=secA−tanA ⇒1-sinAcosA=secA−tanA ⇒secA−tanA=secA−tanA ⇒secA−tanA>0 ⇒secA+tanA>0 ∴secA>0 ⇒A lies in I,IV quadrants

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