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(SinA - 2Sin^2A)/(2Cos^3A-CosA)=tanA |
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Answer» Yes LHS=SinA (1-2sin2A)/cosA(2cos2A-1)=SinA (sin2A+cos2A-2sin2A)/ Cos A (2 cos2A- sin2A- cos2 A)=sinA ( cos2 A- sin2A)/ cos A ( cos2A- sin2A)= sinA/ cos A= tan ALHS=RHS In place of 2 sin^2A We have to take common..... The question us incorrect, i guess. It should be 2 sin^3A It\'s a 2 sin3A |
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