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    Can someone help me starting B. I know that you have to join the two equations in terms of X and y. But how can I use y+1/y. ? Thanks


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    (Original post by Lilly1234567890)
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    Can someone help me starting B. I know that you have to join the two equations in terms of X and y. But how can I use y+1/y. ? Thanks


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    You are given that something is equal to 2sectheta and something else is equal to 2cosectheta... can you think of any (not immediate) relationships between sectheta and costheta?
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    (Original post by Lilly1234567890)
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    Can someone help me starting B. I know that you have to join the two equations in terms of X and y. But how can I use y+1/y. ? Thanks


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    A possibly simpler alternative to what Sean FM is suggesting might be to consider taking the reciprocal of each equation and using a common identity:
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    \sin^2(\theta)+\cos^2(\theta)=1.
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    I'd go with Joostan's method, I'm just going to add a little point that might trip someone up:

    \displaystyle \frac{1}{2\sec \theta} = \frac{1}{x + \frac{1}{x}} = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \neq \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x
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    (Original post by Zacken)
    I'd go with Joostan's method, I'm just going to add a little point that might trip someone up:

    \displaystyle \frac{1}{2\sec \theta} = \frac{1}{x + \frac{1}{x}} = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \neq \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x
    Or
    

\dfrac{4y^2}{y^3 + 1 - 2y^3} = \dfrac{x^3 + 1 - 2x^2}{4x^2}
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    (Original post by Zacken)
    I'd go with Joostan's method, I'm just going to add a little point that might trip someone up:

    \displaystyle \frac{1}{2\sec \theta} = \frac{1}{x + \frac{1}{x}} = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \neq \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x
    thankssssss.
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    thanks allll
    got it
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    (Original post by joostan)
    A possibly simpler alternative to what Sean FM is suggesting might be to consider taking the reciprocal of each equation and using a common identity:
    Spoiler:
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    \sin^2(\theta)+\cos^2(\theta)=1.
    That is exactly what I was suggesting just in a way that gives the OP more room to think for themselves
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    (Original post by SeanFM)
    That is exactly what I was suggesting just in a way that gives the OP more room to think for themselves
    Ah, my bad, but surely this uses the immediate relation between \cos(\theta) and \sec(\theta), which you suggested not to use. . .
 
 
 
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