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Q.

If 1+sinA1−sinA=−(secA+tanA) then the quadrants in which the angle A lies are

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a

I, II

b

II, III

c

I, IV

d

III, IV

answer is B.

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Detailed Solution

1+sinA1−sinA=−(secA+tanA) ⇒1+sinA1+sinA1−sinA1+sinA=−(secA+tanA) ⇒1+sinAcos2A=−(secA+tanA) ⇒cosA is −ve ⇒A∈II,III quadrants

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