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Math Help - Trig Identity Problems

  1. #1
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    Trig Identity Problems

    Got two problems for extra credit that I can't seem to figure out if anyone here can.

    Prove the following.

    1. cosx-(sinx-1)/cosx+(sinx-1) = sinx/1-cosx It says as a hint to multiply
    the numerator and
    denominator of the left side
    by cosx-(sinx-1)

    2. cotx/1-tanx + tanx/1-cotx - 1 = secxcscx

    If anyone happens to know how to do these, please let me know, thanks.
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  2. #2
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    Quote Originally Posted by Chouman View Post
    Got two problems for extra credit that I can't seem to figure out if anyone here can.

    Prove the following.

    1. cosx-(sinx-1)/cosx+(sinx-1) + sinx/1-cosx It says as a hint to multiply
    the numerator and
    denominator of the left side
    by cosx-(sinx-1)

    2. cotx/1-tanx + tanx/1-cotx - 1 = secxcscx

    If anyone happens to know how to do these, please let me know, thanks.

    I already solved the number 1 but i can't solve the number 2...

    Here's the solution:


    cosx - (sinx - 1) / cosx = (sinx - 1) = sinx / 1 - cosx

    *you'll cancel -(sinx - 1) and + (sinx -1) since that the first term is expressed in negative while the other is positive (outside the parentheses)

    cosx / cos x = sinx / 1- cosx

    1 = sinx / sinx

    1=1


    i hope it'll help...
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  3. #3
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    Hello, Chouman!


    1)\;\text{Prove: }\;\frac{\cos x-\sin x +1}{\cos x+ \sin x-1} \:=\:\frac{\sin x}{1-\cos x} . . Ignore their hint

    Multiply top and bottom by (\cos x + \sin x - 1)

    . . \frac{\cos x - \sin x + 1}{\cos x + \sin x - 1}\cdot {\color{blue}\frac{\cos x + \sin x - 1}{\cos x + \sin x - 1}}


    Numerator: . \cos^2\!x + {\color{red}\rlap{/////////}}\sin x\cos x - {\color{green}\rlap{/////}}\cos x - {\color{red}\rlap{/////////}}\sin x\cos x - \sin^2x + \sin x + {\color{green}\rlap{/////}}\cos x + \sin x - 1

    . . . = \;2\sin x - \sin^2\!x - (1 - \cos^2\!x)

    . . . =\;2\sin x - \sin^2\!x - \sin^2\!x

    . . . =\;2\sin x - 2\sin^2\!x

    . . . = \;2\sin x(1 - \sin x)


    Denominator: . \cos^2\!x + \sin x\cos x - \cos x + \sin x\cos x + \sin^2\!x - \sin x - \cos x - \sin x + 1

    . . . = \;\underbrace{\cos^2\!x + \sin^2\!x}_{\text{This is 1}} + 1 - 2\sin x - 2\cos x + 2\sin x\cos x

    . . . = \;2 - 2\sin x - 2\cos x + 2\sin x\cos x

    . . . = \;2(1-\sin x) - 2\cos x(1 - \sin x)

    . . . = \;2(1-\sin x)(1 - \cos x)


    And the fraction becomes: . \frac{{\color{green}\rlap{/}}2\cdot\sin x\cdot({\color{red}\rlap{////////}}1 - \sin x)}{{\color{green}\rlap{/}}2\cdot({\color{red}\rlap{////////}}1-\sin x)\cdot(1 - \cos x)} \;=\;\frac{\sin x}{1 - \cos x} \quad\hdots ta-DAA!

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  4. #4
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    Hello again, Chouman!

    I thinik the second one is even worse ... (Is that possible?)


    2)\;\;\frac{\cot x}{1-\tan x} + \frac{\tan x}{1-\cot x} - 1 \;=\; \sec x\csc x

    The left side is: . \frac{\dfrac{\cos x}{\sin x}} {1 - \dfrac{\sin x}{\cos x}} + \frac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}} - 1


    Multiply top and bottom by \sin x\cos x ... both fractions.

    . . \frac{\sin x\cos x\left(\dfrac{\cos x}{\sin x}\right)} {\sin x\cos x\left(1 - \dfrac{\sin x}{\cos x}\right)} + \frac{\sin x\cos x\left(\dfrac{\sin x}{\cos x}\right)}{\sin x\cos x\left(1-\dfrac{\cos x}{\sin x}\right)} \;-\; 1

    . . =\;\frac{\cos^2\!x}{\sin x\cos x - \sin^2\!x} + \frac{\sin^2\!x}{\sin x\cos x - \cos^2\!x} \:- \:1

    . . = \;\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} + \frac{\sin^2\!x}{\cos x(\sin x - \cos x)} \:-\:1

    . . = \;\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} \:{\color{red}-}\: \frac{\sin^2\!x}{\cos x(\cos x \:{\color{red}-}\: \sin x)} \:=\:1

    . . = \;{\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^2\!x}{\sin x(\cos x - \sin x)} - {\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^2\!x}{\cos x(\cos x - \sin x)} \:-\:1

    . . = \;\frac{\overbrace{\cos^3\!x - \sin^3\!x}^{\text{diff. of cubes}}}{\sin x\cos x(\cos x -\sin x)} \:-\:1

    . . = \;\frac{({\color{red}\rlap{///////////}}\cos x - \sin x)(\cos^2\!x + \sin x\cos x + \sin^2\!x)}{\sin x\cos x({\color{red}\rlap{///////////}}\cos x - \sin x)}\:-\:1

    . . = \;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}} + \sin x\cos x}{\sin x\cos x} \:-\:1 \quad=\quad \frac{1 + \sin x\cos x}{\sin x\cos x} \:-\:1

    . . = \;\frac{1}{\sin x\cos x} + \frac{\sin x\cos x}{\sin x\cos x} \:-\: 1 \quad=\quad \frac{1}{\sin x\cos x} \:+\: 1 \:-\: 1

    . . =\;\frac{1}{\cos x\sin x} \quad=\quad \sec x\csc x \quad\hdots Whew!

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  5. #5
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    Thanks a lot both of you. I appreciate it. The 2nd question most likely is harder because my teacher said they usually get harder as they progress and the 2nd question was after the 1st one.
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