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Math Help - [SOLVED] Proving Identities

  1. #1
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    [SOLVED] Proving Identities

    Hi,
    I am realllly bad at proving identities... so I would appreciate any advice you have, when proving these three identities:

    (4/4a+4) + (7/7a-7) = (2a^2 + 2a/a^2 + 2a + 1) * (a+2)/(a^2+a-2)

    also:

    tan^2x/sec^2x = sin^2x

    also:

    (1+tan^2x)(a-sin^2x)=1

    thank you for any guidance you have to offer
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  2. #2
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    for the 2nd one, so far i have:

    tan^2x/sec^2x = sin^2x

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

    that's all, lol
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  3. #3
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    Hello, ria3!

    The first one is ugly.
    There is no neat way to transform one side to equal the other.

    \frac{4}{4a+4} + \frac{7}{7a-7} \;=\; \frac{2a^2 + 2a}{a^2 + 2a + 1}\cdot\frac{a+2}{a^2+a-2}


    \frac{\tan^2\!x}{\sec^2\!x} \:=\:\sin^2\!x
    The left side is: . \frac{\dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos  ^2\!x}} \;=\;\frac{\sin^2\!x}{\cos^2\!x}\cdot\frac{\cos^2\  !x}{1} \;=\;\sin^2\!x



    (1+\tan^2\!x)(1-\sin^2\!x)\:=\:1
    This one is too easy!

    We have: . \underbrace{(1 + \tan^2\!x)}_{\text{This is }\sec^2\!x}\underbrace{(1 -\sin^2\!x)}_{\text{This is }\cos^2\!x} \;=\;\sec^2\!x\cdot\cos^2\!x \;=\;1

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  4. #4
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    thank you soraban! i really appreciate it.
    i think i've figured out the 1st. it turns out all you have to do is simplify each side by factoring. so, both sides end up being 2a/(a+1)(a-1)
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