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Answers posted by sweetypiesneha18

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answered Jul 30, 2016
2cosx-1=0 => cos2x=1/2 => 2x=cos∏/3 => 2x=(2n+1)∏/3 => x=(2n+1)∏/6
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answered Jul 30, 2016
x=secØ-tanØ=1/cosØ-sinØ/cosØ=1-sinØ/cosØ  y=cosecØ+cotØ=1/sinØ+cosØ/sinØ=1+cosØ/sinØ LHS=
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answered Jul 29, 2016
3[(-cosα)4 +(-sinα)4] - 2[(-cosα)6+(sinα)6] =3[cos4α+sin4α] - 2[cos6α+sin6α] =3[(cos2α+sin2α)2
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