Can you prove this identity?

((1-sint)^2/cos^2t) = (1-sinx/ 1+sinx)

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  • 1 decade ago
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    L.H.S.=(1-sin x)^2/cos^2x

    =(1-sin x)^2/(1-sin^2)

    =(1-sin x)(1-sin x)/(1+sin x)(1-sin x)

    =(1-sin x)/(1+sin x)

    =R.H.S. [Proved]

  • Anonymous
    1 decade ago

    cos^2t = 1 - sin^2t = (1-sint)(1+sint)

    so,

    ( 1- sint)^2 /(1-sint)(1+sint) =(1-sint)/(1+sint)

  • 1 decade ago

    (1 - sin(t))^2/cos^2(t)

    =>[1-sin(t)]^2/(1 - sin^2(t))

    write [1- sin(t)]^2 = [1-sin(t)][1-sin(t)] and

    1 - sin^2(t) = [1 + sin(t)][1 - sin(t)] (since a^2-b^2 = (a+b)(a-b))

    =>[1-sin(t)][1 - sin(t)]/(1+sin(t))(1- sin(t))

    cancel (1- sin(t)) from both numerator and denominator

    =>[1-sin(t)]/[1+ sin(t)]

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