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Math Help - Trig Identities - Simplify and Prove

  1. #1
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    Trig Identities - Simplify and Prove

    I am dying here trying to solve this problem, I'd appreciate if you could help me

    Simplify:
    g(x) = 1/(4sin^2x)(cos^2x) - (1-tan^2x)^2/4tan^2x

    Prove:
    (cotB/cscB-1) + (cotB/cscB+1) = 2secB

    (tan^3x/1+tan^2x) + (cot^3x/1+cot^2x) = 1-2sin^2xcos^2x/sinxcosx

    If you get these your a math god and I will love you for the rest of my life

    ~Thanks

    RedBooster
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  2. #2
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    (cotB/cscB-1) + (cotB/cscB+1) = 2secB

    LS
    [cotB(cscB+1)+cotB(cscB-1)]/[(cscB)^2-1]
    (cotBcscB+cotB+cotBscscB-cotB)/(cotB)^2
    2cotBcscB/(cotB)^2
    2cscB/(cotB)
    2(1/sinB)/(cosB/sinB)
    2(1/sinB)(sinB/cosB)
    2(1/cosB)
    2secB = RS

    I will solve the other ones later
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  3. #3
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    Hello, RedBooster!

    The last one is a killer . . .


    Prove: . \frac{\tan^3\!x}{1+\tan^2\!x} + \frac{\cot^3\!x}{1+\cot^2\!x} \;=\;\frac{1-2\sin^2\!x\cos^2\!x}{\sin x\cos x}

    The left side is: . \frac{\dfrac{\sin^3\!x}{\cos^3\!x}} {1 + \dfrac{\sin^2\!x}{\cos^2\!x}} + \frac{\dfrac{\cos^3\!x}{\sin^3\!x}} {1 + \dfrac{\cos^2\!x}{\sin^2\!x}}

    . . = \; \frac{{\color{blue}\cos^3\!x}\left(\dfrac{\sin^3\!  x}{\cos^3\!x}\right)} {{\color{blue}\cos^3\!x}\left(1 + \dfrac{\sin^2\!x}{\cos^2\!x}\right)} + \frac{{\color{blue}\sin^3\!x}\left(\dfrac{\cos^3\!  x}{\sin^3\!x}\right)}  {{\color{blue}\sin^3\!x}\left(1 + \dfrac{\cos^2\!x}{\sin^2\!x}\right)}

    . . = \;\frac{\sin^3\!x}{\cos^3\!x+\sin^2\!x\cos x} + \frac{\cos^3\!x}{\sin^3\!x + \sin x\cos^2\!x}

    . . = \;\frac{\sin^3\!x}{\cos x\underbrace{(\sin^2\!x+\cos^2\!x)}_{\text{This is 1}}} + \frac{\cos^3\!x}{\sin x\underbrace{(\sin^2\!x+\cos^2\!x)}_{\text{This is 1}}}

    . . = \;\frac{\sin^3\!x}{\cos x} + \frac{\cos^3\!x}{\sin x}

    . . =\;{\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^3\!x}{\cos x} + {\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^3\!x}{\sin x}

    . . =\;\frac{\sin^4\!x + \cos^4\!x}{\sin x\cos x}

    . . =\;\frac{\sin^4\!x \:{\color{red}+ \:2\sin^2\!x\cos^2\!x} + \cos^4\!x \:{\color{red}- \:2\sin^2\!x\cos^2\!x}}{\sin x\cos x}

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

    . . = \;\frac{1 - 2\sin^2\!x\cos^2\!x}{\sin x\cos x} \quad\hdots . ta-DAA!

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  4. #4
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    Wow, thank you soo much! This really has helped me
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