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Prove the following identities sec(x) - sin(x) = [tan^2(x) + cos^2(x)] / [sec(x) + sin(x)]?

sec(x) - sin(x) = [tan^2(x) + cos^2(x)] / [sec(x) + sin(x)]

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  • ?
    Lv 4
    1 decade ago
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    sec(x) - sin(x) = [tan^2(x) + cos^2(x)] / [sec(x) + sin(x)]

    LS:

    sec(x) - sin(x)

    = [sec(x) - sin(x)]*[sec(x) + sin(x)] / [sec(x) + sin(x)]

    =[sec^2(x) - sin^2(x)] / [sec(x) + sin(x)]

    now must show:

    sec^2(x) - sin^2(x) = tan^2(x) + cos^2(x)

    LS:

    sec^2(x) - sin^2(x)

    =sec^2(x) - 1 + cos^2(x)

    aside:

    sin^2(x) + cos^2(x) = 1 dividing through by cos^2(x)

    tan^2(x) + 1 = sec^2(x)

    tan^2(x) = sec^2(x) - 1

    resubstituting tan^2(x):

    =sec^2(x) - 1 + cos^2(x)

    = tan^2(x) + cos^2(x) as required.

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  • 1 decade ago

    R. H. S.=[tan^2 x + cos^2 x] / [sec x + sin x]

    =[sec^2 x - 1 +cos^2 x] / [sec x + sin x] [since tan^2x=sec^2x - 1]

    =[sec^2 x - (1 - cos^2 x)] / [sec x + sin x]

    =[sec^2 x - sin^2 x] / [sec x+ sin x]

    =[sec x + sin x][sec x - sin x] / [sec x + sin x] [since a^2 - b^2=(a+b)(a-b)]

    =sec x - sin x [cancelling (sec x + sin x) in both numerator n denominator]

    =L. H. S.

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