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    I've been stuck on this question for ages, no matter what I do I can't get the right answer.
    Using the identities tan(x)=sin(x)/cos(x) and cos^2(x)+sin^2(x)=1, prove that tan(x)sin(x)/1-cos(x) is equal to 1 + 1/cos(x).

    If anyone has the AQA Pure core Maths 1 and 2 textbook, it's page 298, question 3j. Thanks in advance for any help!
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    Multiply the left hand side by \displaystyle \frac{1+\cos x}{1+ \cos x}
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    (Original post by King Hotpie)
    I've been stuck on this question for ages, no matter what I do I can't get the right answer.
    Using the identities tan(x)=sin(x)/cos(x) and cos^2(x)+sin^2(x)=1, prove that tan(x)sin(x)/1-cos(x) is equal to 1 + 1/cos(x).

    If anyone has the AQA Pure core Maths 1 and 2 textbook, it's page 298, question 3j. Thanks in advance for any help!
    start by replacing tan x with sin x/cos x
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    (Original post by King Hotpie)
    I've been stuck on this question for ages, no matter what I do I can't get the right answer.
    Using the identities tan(x)=sin(x)/cos(x) and cos^2(x)+sin^2(x)=1, prove that tan(x)sin(x)/1-cos(x) is equal to 1 + 1/cos(x).

    If anyone has the AQA Pure core Maths 1 and 2 textbook, it's page 298, question 3j. Thanks in advance for any help!
    Multiply top and bottom of [tan(x)sin(x)]/[1-cos(x)] by '1+cos(x)'; then use the given identities.

    EDIT: Far too slow.
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    (Original post by Mr M)
    Multiply the left hand side by \displaystyle \frac{1+\cos x}{1+ \cos x}

    (Original post by Farhan.Hanif93)
    Multiply top and bottom of [tan(x)sin(x)]/[1-cos(x)] by '1+cos(x)'; then use the given identities.

    EDIT: Far too slow.
    Pfft..you guys are ganging up on me!! It works OK if you just replace tan, then replace sin^2, then rewrite top and factorize as difference of 2 squares
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    edit: nevermind
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    (Original post by Aquar)
    Using the identities \displaystyle \tan{(x)}=\frac{sin{(x)}}{\cos{(  x)}} and \displaystyle \cos^{2}{(x)}+\sin^{2}{(x)}=1, prove that \displaystyle\frac{\tan{(x)}\sin  {(x)}}{1-\cos{(x)}}=1+\frac{1}{\cos{(x)}}.

    Is this what you mean? I don't have the textbook.
    Yes. I'm afraid several people answered the question while you were busy LaTexing!
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    (Original post by Mr M)
    Yes. I'm afraid several people answered the question while you were busy LaTexing!
    I genuinely didn't understand the question, I had to see it set out and hopefully now I can have a go at this.
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    Got it! Thank you all very much for your help, it seems so annoyingly simple now I've finally done it... Oh well, onwards and upwards with the rest of the questions!
 
 
 
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