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    Are there any solutions for cotx=0, because that means 1/tanx=0, and rearranging will give 1=0, so I didn't think this was possible. But the graph of cotx intercepts the x axis, so im kind of confused.
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    cotx = 1/tanx = cosx/sinx = 0...
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    cotx = cosx/sinx so if cotx = 0, cosx must =0

    arccos 0 = pi/2 etc.
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    Replace cotx with cosx/sinx.
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    Yes, wherever tan(x) is infinity, then cot(x)=0. But we're not allowed to discuss the exam....
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    (Original post by zafaru123)
    Are there any solutions for cotx=0, because that means 1/tanx=0, and rearranging will give 1=0, so I didn't think this was possible. But the graph of cotx intercepts the x axis, so im kind of confused.
    I think that's a false move. You can't move 0 down, coz in effect you are dividing by 0 on both sides. And dividing by 0 is not allowed.
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    1/tanx=0

    tanx=0 .....
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    Cotx= 1/tanx

    = 1/ (sinx/cosx)
    =cosx/sinx = 0

    => cos x = 0.
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    Im wondering about this aswell, on the cot graph, it does cross the x axis, but if you look at it in terms of tan, its impossible.
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    You must be drawing the graph wrong. http://www4c.wolframalpha.com/Calcul...image/gif&s=22

    I think you may also be looking at x=0 which does indeed have no value.
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    I've done this in a past paper.
    If cotx = 0, then tanx ---> infinity. (tanx tends to infinity)
    So, the solutions will be 90, 270 etc. (wherever the tan graph lines go to asymptotes).
 
 
 
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Updated: January 23, 2011
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