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    Hi, I need some help with this...

    Prove that:
    1 + [1 / tan^2 A]

    is equal to

    1 / [sin^2 A]



    I hope that makes sense. The / means divide and ^2 means to the power of 2 obviously.

    Thanks!
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    Write 1 as tan^2 A / tan^2 A.
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    The way to learn these is to mess about a lot , getting them wrong even, and trying different things, until you see what sort of methods work.
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    (Original post by generalebriety)
    Write 1 as tan^2 A / tan^2 A.
    Is it not better to write the 1 as Sin^2 / Sin^2 ?
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    (Original post by generalebriety)
    Write 1 as tan^2 A / tan^2 A.
    Then what? :p:
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    (Original post by Narik)
    Then what? :p:
    Your thread title is "trigonometric identities". Why don't you try using a trigonometric identity? :confused:
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    (Original post by stevencarrwork)
    Is it not better to write the 1 as Sin^2 / Sin^2 ?
    Well, maybe (though I don't really know what you mean by "better" other than that the identities involved are more well known), but it's a lot less obvious. There's no point in giving a method that people would never spot in the exam.
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    (Original post by generalebriety)
    Your thread title is "trigonometric identities". Why don't you try using a trigonometric identity? :confused:
    Tan = sin/ cos is one identity.

    Sin^2 + Cos^2 is another.

    Bash them in and see what happens.
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    (Original post by stevencarrwork)
    Tan = sin/ cos is one identity.

    Sin^2 + Cos^2 is another.

    Bash them in and see what happens.
    I think you need to read the posting guidelines. I know how to do the problem, and so do you; I was trying to lead the OP towards finding their own solution, not to hand it to them on a plate.
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    (Original post by generalebriety)
    I think you need to read the posting guidelines. I know how to do the problem, and so do you; I was trying to lead the OP towards finding their own solution, not to hand it to them on a plate.
    Plate please!
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    (Original post by Narik)
    Plate please!
    Sounds like you need to read the posting guidelines too.
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    (Original post by Narik)
    Plate please!
    Try substitution, see if that works. I doubt anyone will give it you on a plate; otherwise you'll be straight back here as soon as you see another one of the same question.
 
 
 
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