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Question

lf 1+sinA1sinA=secA+tanA, then A Iies in the quadrants

A
I, II
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B
II, III
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C
I, III
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D
I, IV
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Solution

The correct option is D I, IV
1+sinA1sinA=(1+sinA)2(1+sinA)(1sinA)
=(1+sinA)2cos2A
=(secA+tanA)2
=|secA+tanA|
=secA+tanA (as given)
let secA+tanA=x
so we get |x|=x
this is possible only when x 0
so secA+tanA0
1+sinAcosA0
this means that cosA>0
this is possible only in first and fourth quadrants

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