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    Please help me solve 9iii, Attachment 729964
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    (Original post by joyoustele)
    Please help me solve 9iii, Attachment 729960
    not able to see it...

    could you post it in imagepost.org and link from there ?
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    (Original post by joyoustele)
    Please help me solve 9iii,
    Just use https://imgur.com/ or something to upload images then link them here. By now it should be obvious that TSR's image upload procedure is terrible.
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    (Original post by the bear)
    not able to see it...

    could you post it in imagepost.org and link from there ?
    https://imgur.com/a/gHTjd
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    (Original post by RDKGames)
    Just use https://imgur.com/ or something to upload images then link them here. By now it should be obvious that TSR's image upload procedure is terrible.
    https://imgur.com/a/gHTjd
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    (Original post by joyoustele)
    Please help me solve 9iii,
    For 9iii you need to use the identity in part (i) and obtain a quadratic in \tan \theta while k^2 would just be some constant. From there, you want to show that b^2 - 4ac > 0. This is C1 stuff, except you got tan as the variable.
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    (Original post by RDKGames)
    For 9iii you need to use the identity in part (i) and obtain a quadratic in \tan \theta while k^2 would just be some constant. From there, you want to show that b^2 - 4ac > 0. This is C1 stuff, except you got tan as the variable.
    Ohhh...Thanks
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    (Original post by RDKGames)
    For 9iii you need to use the identity in part (i) and obtain a quadratic in \tan \theta while k^2 would just be some constant. From there, you want to show that b^2 - 4ac > 0. This is C1 stuff, except you got tan as the variable.
    That doesnt work, here is what the Answer says, However I dont understand the bit in red. (I got the equation)

    Name:  9iii answer.JPG
Views: 12
Size:  25.1 KB
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    (Original post by joyoustele)
    That doesnt work, here is what the Answer says, However I dont understand the bit in red. (I got the equation)
    It does work it's just a different approach and requires a bit of alternative thinking, though I did misread the question initially so this isn't the best approach.

    As for the red bit, the RHS is always positive - this should be obvious - and so the root of that will yield that \tan \theta can be strictly +ve or strictly -ve. In either case, there will be a single solution for each (consider the graph of \tan \theta for 0 < \theta < \pi, which coveres the +ve and -ve values exactly once.)
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    (Original post by RDKGames)
    It does work it's just a different approach and requires a bit of alternative thinking, though I did misread the question initially so this isn't the best approach.

    As for the red bit, the RHS is always positive - this should be obvious - and so the root of that will yield that \tan \theta can be strictly +ve or strictly -ve. In either case, there will be a single solution for each (consider the graph of \tan \theta for 0 < \theta < \pi, which coveres the +ve and -ve values exactly once.)
    What is a quadrant
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    (Original post by joyoustele)
    What is a quadrant
    It's probably referring to CAST diagram. I don't use them so I can't say.
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    (Original post by RDKGames)
    It's probably referring to CAST diagram. I don't use them so I can't say.
    alright thanks
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    (Original post by joyoustele)
    What is a quadrant
    https://dryuc24b85zbr.cloudfront.net...=1493912512633
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    Lol, you've got the poundland picture as your profile
 
 
 
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