Question
Sun March 11, 2012 By: Mahesh Joshi

problem

Expert Reply
Sun March 11, 2012
Taking LHS
(sinA-1/sinA)(cosA-1/cosA)(sinA/cosA + cosA/sinA)
= (sin2A - 1)/sinA . (cos2A-1)/cosA . (sin2A + cos2A)/sinAcosA
= -cos2A/sinA . -sin2A/cosA . 1/sinAcosA = 1 = RHS
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