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(SinA - 2Sin^2A)/(2Cos^3A-CosA)=tanA

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(SinA - 2Sin^2A)/(2Cos^3A-CosA)=tanA

    • Posted by Deep Rai 7 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 5 answers

    Nancy Rajput 7 years, 6 months ago

    Yes

    0Thank You

    Nancy Rajput 7 years, 6 months ago

    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 A LHS=RHS

    0Thank You

    Afghhhh Ggfghhh 7 years, 6 months ago

    In place of 2 sin^2A

    0Thank You

    Afghhhh Ggfghhh 7 years, 6 months ago

    We have to take common..... The question us incorrect, i guess. It should be 2 sin^3A

    0Thank You

    Nancy Rajput 7 years, 6 months ago

    It's a 2 sin3A

    0Thank You

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