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Math Help - Trig Identities - where is my error?

  1. #1
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    Trig Identities - where is my error?

    Prove:
    csc(A)^2 + sec(A)^2 = csc(A)^2sec(A)^2

    left side is:
    (1/cos(A)^2) + (1/sin(A)^2)

    I multiply the first term by (sin(A)^2)/(sin(A)^2)
    and the second term by (cos(A)^2)/(cos(A)^2)

    (sin(A)^2)/(sin(A)^2)(1/cos(A)^2) + (cos(A)^2)/(cos(A)^2)(1/sin(A)^2)

    Now the two terms have a common denominator and can be combined and we have:
    (sin(A)^2 + cos(A)^2)/(2cos(A)^2sin(A)^2)

    Then numerator = 1:
    1/(2cos(A)^2sin(A)^2)

    Then we have
    (1/2)csc(A)^2sec(A)^2

    Where does the 1/2 come from? What am I doing wrong? How can I make the left side look like csc(A)^2sec(A)^2 ???

    Thanx

    -Tony
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  2. #2
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    Grandad's Avatar
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    Hello Tony

    Welcome to Math Help Forum!
    Quote Originally Posted by Xplorer View Post
    ...

    (sin(A)^2)/(sin(A)^2)(1/cos(A)^2) + (cos(A)^2)/(cos(A)^2)(1/sin(A)^2)

    Now the two terms have a common denominator and can be combined and we have:
    (sin(A)^2 + cos(A)^2)/(2cos(A)^2sin(A)^2)
    Ouch! Since when have you added fractions like that?

    \frac{a}{b}+\frac{c}{b}=\frac{a+c}{b},
    not \frac{a+c}{2b}.

    Grandad
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  3. #3
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    Oh my gosh!!! Now I see it. It was late at night when I was doing that. Wow I feel stupid now

    Thank you very much. lol!

    -Tony
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