lf √1+sinA1−sinA=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+sinA1−sinA=√(1+sinA)2(1+sinA)(1−sinA) =√(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+tanA⩾0
1+sinAcosA≥0
this means that cosA>0 this is possible only in first and fourth quadrants