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need help proving identity?

sin θ 1+cos θ

------------- +-------------- = 2cot θsecθ

1+cos θ sin θ

3 Answers

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  • M3
    Lv 7
    1 decade ago
    Favorite Answer

    θ has been left out for greater clarity in reading

    ▬▬▬▬▬▬▬▬▬ ▬▬▬▬▬▬▬▬▬▬▬

    LHS = sin / (1+cos) + (1+cos) / sin

    put LHS under a common denominator

    [sin² + (1+cos)² ] / (1+cos)•sin

    (sin² + 1 + cos² + 2cos) / (1+cos)•sin

    2(1+cos) / (1+cos)•sin

    2/sin

    multiply top & bottom by cos

    2cos/(sin•cos)

    2cot•sec

    QED

  • 1 decade ago

    from LHS

    sin x / (1 + cos x) + (1 + cos x) / sin x

    =(sin^2 x + 1 + 2 cos x + cos^2 x ) / (sin x)( 1 + cos x)

    =( 2 + 2 cos x ) / (sin x)( 1 + cos x)

    = 2 / sin x

    = 2 cos x / (sin x cos x)

    = 2cot x sec x

    for more info check out http://www.a-maths-tuition.com/2010/02/proving-tri...

  • Mrs.W
    Lv 4
    1 decade ago

    we will work the LHS going for a common denominator of sinx*(1+cosx) .

    [The restrictions are that the cos x does not equal -1, the cos x does not equal 0, the sinx does not equal zero]

    [sin^2 (x) +1+2cosx + cos^2 (x) ]/ sinx(1+cosx)

    [2 + 2cosx] / sinx* (1+cosx)

    2(1 + cosx) / sinx (1+cosx)

    2/ sinx

    ==>multiply by cosx/cosx

    2cosx/sinx cosx

    ===> split apart into the product of 3 terms

    2 * cosx/sinx * 1/ cosx

    2 * cotx * sec x = 2 *cotx* secx

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