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Math Help - trig id

  1. #1
    Member pberardi's Avatar
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    trig id

    Hey all, I was wondering if someone can walk me through this one:

    [cot^2(B) - cos^2(B)]/[csc^2(B) -1]

    My solution is as follows:
    [cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

    -cos^4(B)/sin^2(B) +1

    and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

    Thanks,
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by pberardi View Post
    Hey all, I was wondering if someone can walk me through this one:

    [cot^2(B) - cos^2(B)]/[csc^2(B) -1]

    My solution is as follows:
    [cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

    -cos^4(B)/sin^2(B) +1

    and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

    Thanks,
    From where you got to:

    Distribute the minus sign to the denominator:

    \frac{cos^4(B)}{-(sin^2(B)+1)} = \frac{cos^4(B)}{1-sin^2(B)}

    and what does 1-sin^2(B) equal
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  3. #3
    Junior Member JoanF's Avatar
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    Quote Originally Posted by pberardi View Post
    Hey all, I was wondering if someone can walk me through this one:

    [cot^2(B) - cos^2(B)]/[csc^2(B) -1]

    My solution is as follows:
    [cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

    -cos^4(B)/sin^2(B) +1

    and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

    Thanks,
    Hi!

    I think you did wrong on the signs. Because you should get -cos^4(B)/sin^2(B) - 1. If you get this, you get the exercice correct:



    If you want to see my resolution, here it is:

    Last edited by JoanF; April 17th 2009 at 05:13 PM.
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  4. #4
    Junior Member JoanF's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    From where you got to:

    Distribute the minus sign to the denominator:

    \frac{cos^4(B)}{-(sin^2(B)+1)} = \frac{cos^4(B)}{1-sin^2(B)}

    and what does 1-sin^2(B) equal

    But
    \frac{cos^4(B)}{-(sin^2(B)+1)} =\frac{cos^4(B)}{-sin^2(B)-1}
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  5. #5
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    Quote Originally Posted by pberardi View Post
    Hey all, I was wondering if someone can walk me through this one:

    [cot^2(B) - cos^2(B)]/[csc^2(B) -1]

    My solution is as follows:
    [cot^2(B)/-sec^2(B)] - cos^2(B)/-sec^2(B)

    -cos^4(B)/sin^2(B) +1

    and now i am a bit stuck. the correct answer is cos^2(B). Can someone please finish this off or tell me a better way to get to the right answer?

    Thanks,
    Here is a better way, see,

    \frac{\cot^2 B-\cos^2 B}{\csc^2 B-1}

    =\frac{\cot^2 B-\cos^2 B}{\cot^2 B}

    =\frac{\cot^2 B}{\cot^2 B}-\frac{\cos^2 B}{\cot^2 B}

    =1-\frac{\cos^2 B}{\cot^2 B}

    =1-\frac{\cos^2 B}{\frac{\cos^2 B}{\sin^2 B}}

    =1-\cos^2 B\times \frac{\sin^2 B}{\cos^2 B}

    =1-\sin^2 B

    =\cos^2 B
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