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If 0<α<π then 1sinα+cot2α−cos2α

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a

sinα

b

−sinα

c

1sin α

d

1

answer is A.

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Detailed Solution

α∈(0,π) then α∈I,II quadrants, sinα is always positive in first and second quadrant1sinα+cot2α−cos2α=1sinα+cos2α(cot2α) =1sinα+cos2αsinα = sinα

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