Trig Identities help

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    Hey guys,

    I'm a little stuck as to how to complete this.

    Name:  ImageUploadedByStudent Room1475805833.092003.jpg
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    I've gotten up to (1-cos^2x)/cos^2 - 1 but can't seem to work out the rest


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    Try working with the LHS to produce something that you can manipulate using the \cos^2x
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    (Original post by TheBBQ)
    Try working with the LHS to produce something that you can manipulate using the \cos^2x
    I have no idea what that's meant to mean. (Self-teaching noob to maths)


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    (Original post by nemanuel96)
    I have no idea what that's meant to mean. (Self-teaching noob to maths)


    Posted from TSR Mobile
    LHS = Left hand side of the equation.
    See what you can do with that first
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    (Original post by TheBBQ)
    LHS = Left hand side of the equation.
    See what you can do with that first
    Haha I know what that means. The mar scheme's next step is:

    1/cos^2x - cos^2x/cos^2x - 1

    But I'm not entirely sure how they've managed to do this. ☹️


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    (Original post by nemanuel96)
    Hey guys,

    I'm a little stuck as to how to complete this.

    Name:  ImageUploadedByStudent Room1475805833.092003.jpg
Views: 25
Size:  41.4 KB

    I've gotten up to (1-cos^2x)/cos^2 - 1 but can't seem to work out the rest


    Posted from TSR Mobile
    Just a hint,

     \frac{x}{2} + \frac{3}{2} = \frac{x+3}{2}


    So use this in reverse
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    (Original post by asinghj)
    Just a hint,

     \frac{x}{2} + \frac{3}{2} = \frac{x+3}{2}


    So use this in reverse
    Thanks a lot! It makes much more sense now


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    I've gotten up to this question:

    Name:  ImageUploadedByStudent Room1475851131.470750.jpg
Views: 16
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    I've so far managed to get up to sinx/cosx + cosx/sinx

    The mark scheme's next steps:

    Name:  ImageUploadedByStudent Room1475851255.361181.jpg
Views: 16
Size:  27.6 KB

    I don't understand how they've transitioned from what I've gotten up to, to getting sinx^2 + cosx^2/sinxcosx. Did they square both numerator and denominator to get to there?




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    (Original post by nemanuel96)
    I've gotten up to this question:

    Name:  ImageUploadedByStudent Room1475851131.470750.jpg
Views: 16
Size:  11.0 KB

    I've so far managed to get up to sinx/cosx + cosx/sinx

    The mark scheme's next steps:

    Name:  ImageUploadedByStudent Room1475851255.361181.jpg
Views: 16
Size:  27.6 KB

    I don't understand how they've transitioned from what I've gotten up to, to getting sinx^2 + cosx^2/sinxcosx. Did they square both numerator and denominator to get to there?

    * * *Posted from TSR Mobile
    they just applied the method of adding fractions a/b + c/d = ( ad + bc *)/bd
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    (Original post by the bear)
    they just applied the method of adding fractions a/b + c/d = ( ad + bc *)/bd
    Thank you so much. Haha, my maths is a little rusty.


    Posted from TSR Mobile
 
 
 
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