Trig problems need help

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    How can I shows that these two expressions underlined are the same ? Name:  1475087049668952323006.jpg
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    (Original post by coconut64)
    How can I shows that these two expressions underlined are the same ? Name:  1475087049668952323006.jpg
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    You seem to have some confusion about fractions here...

    Step 1) Note  \displaystyle \frac{1}{abc} = \frac{1}{a} \times \frac{1}{b} \times \frac{1}{c}

    Step 2) What's the definition of  \displaystyle \sec(x) ? What about  \displaystyle \cot(x) ?

    Piece them together.
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    (Original post by Blazy)
    You seem to have some confusion about fractions here...

    Step 1) Note  \displaystyle \frac{1}{abc} = \frac{1}{a} \times \frac{1}{b} \times \frac{1}{c}

    Step 2) What's the definition of  \displaystyle \sec(x) ? What about  \displaystyle \cot(x) ?

    Piece them together.
    I have managed to work it out from one side but not the other ...Name:  1475090085804952323006.jpg
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    cos2y is 1/sec2y, like you said on the left. Just apply that to the RHS as well and you should be sorted - don't just change the cot term
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    Well, you have to recall that
    sec2x=1/cos2x and
    cot2x=1/tan2x
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    (Original post by bstone13)
    cos2y is 1/sec2y, like you said on the left. Just apply that to the RHS as well and you should be sorted - don't just change the cot term
    how can I get cos2y at the bottom? I only have cos^2 2y at the top of my fraction and this wouldn't give me sec2y... Thanks
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    (Original post by coconut64)
    how can I get cos2y at the bottom? I only have cos^2 2y at the top of my fraction and this wouldn't give me sec2y... Thanks
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    You have to go back to the beginning. You didn't sub in for cos2y when you should have.

    Does it make sense now?
 
 
 
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