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$I f n b e t h e n u m b e r o f s o l u t i o n s o f t h e e q u a t i o n |\cot x| = \cot x + \frac{1}{\sin x} (0 < x < 2 π), t h e n n =$

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Correct option is A

cotx=−cotx, cotx<0cotx, cotx≥0Case(i): Let cotx≥0. Then cotx=cotx+1sinx 1sinx=0 sinx=∞ which is impossible Case(ii): Let cotx<0. Then -cotx=cotx+1sinx ⇒cosx=-12 ⇒x=2π3,4π3 ∵0

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If n be the number of solutions of the equation cotx=cotx+1sinx0<x<2π, then n=

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$I f n b e t h e n u m b e r o f s o l u t i o n s o f t h e e q u a t i o n |\cot x| = \cot x + \frac{1}{\sin x} (0 < x < 2 π), t h e n n =$