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March 28, 2017

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sin(theta)/1-cos(theta) + 1-cos(theta)/sin(theta) = 2csc(theta)

That question makes absolutely no sense.. could someone help me? or lead me in the direction to figuring it out?

and..
(Beside the trig functions is theta)

1+1/cos = tan^2/sec-1

  • precalc h - ,

    maybe you mean
    sin t/(1-cos t) + (1-cos t)/sin t

    sin^2 t (1-cos t) (1-cos t)^2
    ----------------- + ----------------
    (1-cos t)(sin t) ______ (1-cos t)(sin t)

    numerator only for a while

    sin^2 t-sin^2 t cos t +1 -2 cos t+cos^2 t
    but sin^2 + cos^2 = 1
    2 -sin^2 t cos t - 2 cos t
    put denominator back
    2 -sin^2 t cos t - 2 cos t
    --------------------------
    (1-cos t)(sin t)

    I think I misunderstood your lack of parentheses but I think you can get the idea
    BE CAREFUL ABOUT PARENTHESES!!
    when typing on the computer i can not tell what you mean for numerators or denominators.

  • error fx - ,

    sin t/(1-cos t) + (1-cos t)/sin t

    sin^2 t ______________ (1-cos t)^2
    ----------------- + ----------------
    (1-cos t)(sin t) ______ (1-cos t)(sin t)

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

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

    = 2/sin t
    which is csc t

  • precalc h - ,

    The parenthesis were for the arguments.. My question was how do I verify this indentity

  • precalc h - ,

    Theta = argument..

  • precalc h - ,

    I did verify it.
    I only did the left side, starting with what you gave me and ending with
    2/sin t
    which is 2 csc t
    which is the right side

  • precalc h - ,

    Here is how you should have written it
    (use t for theta)

    sin(theta)/ { 1-cos(theta)} + {1-cos(theta)}/sin(theta) = 2 csc(theta)

  • precalc h - ,

    The critical point is that afteter you get it all over the LCD (sin t)(1-cos t)
    you see that you have sin^2 t+ cos^2 t which is one in the numerator.

  • precalc h - ,

    I don't understand what you did?

  • precalc h - ,

    How would x'ing the right side of the plus side by sin give you (1-cos)^2

  • precalc h - ,

    OH nevermind about that .

  • precalc h - ,

    okay I get it now, thanks so much. I have another one at the bottom of this question.. could you help me with it? But THANKYOU SO MUCHCHCH! I get it. :--)

  • precalc h - ,

    Only work on the left side of your equal sign

    Your LCD on the left is
    (1-cos t)(sin t)

    to get that on the bottom of the first term you multiply top and bottom by sin t

    to get it on the bottom of the second term you multiply top and bottom by (1 -cos t)

    when you multiply (1-cos t)times (1-cos t) you get (1-cos t)^2

  • precalc h - ,

    1+1/cos = tan^2/sec-1

    I will guess (parentheses missing again) you mean

    1+1/cos = tan^2/(sec-1 )

    use sin and cos everywhere

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

    multiply top and bottom of right by cos^2

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

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

    (cos + 1)/cos = (1-cos^2)/cos(1-cos)
    but (1-cos^2)= (1+cos)(1-cos)

    (cos + 1)/cos = (1+cos)(1-cos)/cos(1-cos)

    (cos + 1)/cos = (1+cos)/cos

  • precalc h - ,

    dank memes

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