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Math Help - Trig. Problem

  1. #1
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    Angry Trig. Problem

    Please help solving (prove) following equation-

    1+tan^2A/1+cot^2A=(1-tanA/1-cotA)^2

    please read above as tan square A & cot square A
    Thanks
    Last edited by raj01b; August 25th 2009 at 04:31 AM. Reason: not the equation I am stuck with
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  2. #2
    MHF Contributor red_dog's Avatar
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    It's not an equation, it's an identity.

    The left side member:

    \frac{1+\tan^2A}{1+\cot^2A}=\frac{1+\tan^2A}{1+\fr  ac{1}{\tan^2A}}=\frac{1+\tan^2A}{\frac{1+\tan^2A}{  \tan^2A}}=\tan^2A

    \left(\frac{1+\tan A}{1+\cot A}\right)^2=\left(\frac{1+\tan A}{1+\frac{1}{\tan A}}\right)^2=

    =\left(\frac{1+\tan A}{\frac{1+\tan A}{\tan A}}\right)^2=\tan^2A

    Then the identity holds.
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  3. #3
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    Hello, raj01b!

    Another approach . . .


    Prove: . \frac{1+\tan^2\!A}{1+\cot^2\!A} \;=\;\left(\frac{1+\tan A}{1+\cot A}\right)^2
    The left side is: . \frac{1 + \tan^2\!A}{1 + \cos^2\!A} \;=\;\frac{\sec^2\!A}{\csc^2\!A} \;=\; \frac{\dfrac{1}{\cos^2\!A}}{\dfrac{1}{\sin^2\!A}} \;=\;\frac{\sin^2\!A}{\cos^2\!A} \;=\;\tan^2\!A

    Multiply by \frac{(1+\tan A)^2}{(1 +\tan A)^2}\!:\quad\tan^2\!A\cdot\frac{(1+\tan A)^2}{(1+\tan A)^2} \;=\;\left[\frac{\tan A(1 + \tan A)}{1 + \tan A}\right]^2


    Divide top and bottom by \tan A\!:\quad\left(\frac{1+\tan A}{\frac{1}{\tan A} + 1}\right)^2  \;=\;\left(\frac{1+\tan A}{1 + \cot A}\right)^2

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  4. #4
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    trig problem

    Thanks for the help but equation I posted earlier was not the equation I was stuck with, actually following is the equation I want to solve-

    1+tan^2 A/1+cot^2 A= (1-tan A/1-cot A)^2
    sorry for the trouble, any further help will be appreciated.

    Thanks
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  5. #5
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    Hello raj01b
    Quote Originally Posted by raj01b View Post
    Thanks for the help but equation I posted earlier was not the equation I was stuck with, actually following is the equation I want to solve-

    1+tan^2 A/1+cot^2 A= (1-tan A/1-cot A)^2
    sorry for the trouble, any further help will be appreciated.

    Thanks
    You've already seen how \frac{1+\tan^2A}{1+\cot^2A}= \tan^2A

    Well if we multiply \Big(\frac{1-\tan A}{1-\cot A}\Big)^2 'top-and-bottom' by \tan^2A, we get:

    \Big(\frac{1-\tan A}{1-\cot A}\Big)^2=\Big(\frac{1-\tan A}{1-\cot A}\Big)^2\times \frac{\tan^2A}{\tan^2A}

    = \frac{\tan^2A(1-\tan A)^2}{(\tan A - \cot A\tan A)^2}

    = \frac{\tan^2A(1-\tan A)^2}{(\tan A - 1)^2}

    = \tan^2A, since (1-\tan A)^2 = (\tan A -1)^2

    So \frac{1+\tan^2A}{1+\cot^2A}=\Big(\frac{1-\tan A}{1-\cot A}\Big)^2

    Grandad
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  6. #6
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    trig problem

    Thanks Soroban for the quick reply. I was actually confused that (1-tan A)^2 & (tan A-1)^2 are the same thing.

    Thank you very much once again.
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