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    Basically I have no idea whatsoever how to answer this question:

    prove that cot (A+B) = cot A cot B -1 / cot A + cot B


    the answer is in the back of the book but i would love to know how to solve this so i can do it on my own next time. please help someone!
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    tan x = sin x / cos x
    so cot x = cos x / sin x
    start here
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    What do you know about the identity for  tan (A+B) and the relationship between  tan x and  cot x ?
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    (Original post by Goldfishy)
    What do you know about the identity for  tan (A+B) and the relationship between  tan x and  cot x ?
    cot x is 1/tan maybe??
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    @IAmTheChosenOne you could do it like that but that would be long - just use the addition formula and the fact that the usual relation between tan and cot
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    i would start by swapping cot by 1/tan then addition formula, then once simplified change back to cot
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    even if i did replace cot (a+b) with 1/tan i would still be clueless :0
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    (Original post by scholarshipkid)
    cot x is 1/tan maybe??
    Yep - don't hesitate :yep: You can apply this to the the tan addition formula.

    If you do manage that, you might like to investigate similar identities. such as  sec (A+B) or  cosec (A+B) to prove that you know how to do it.
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    so

    1/tan(a+b)

    where would i go from here?
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    it confuses me because it's a fraction!!
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    okay so now i've got:

    tan a + tan b / 1-tan a x tan b

    stuckkkk!
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    (Original post by scholarshipkid)
    even if i did replace cot (a+b) with 1/tan i would still be clueless :0
    If  tan (A+B) = \frac {x}{y}
    then  cot (A+B) = \frac {1}{tan (A+B)} = \left(\frac{x}{y}\right)^{-1} = \frac {y}{x}
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    (Original post by Goldfishy)
    If  tan (A+B) = \frac {x}{y}
    then  cot (A+B) = \frac {1}{tan (A+B)} = \left(\frac{x}{y}\right)^{-1} = \frac {y}{x}

    does cot a + b flip the fraction then?
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    (Original post by scholarshipkid)
    does cot a + b flip the fraction then?
    Yes because cot is the reciprocal of tan.

    But you don't need to use this - you can prove it directly from the fact that cot X = cos X / sin X, so cot (A + B) = cos(A + B) / sin (A + B) and then use the usual sum formulae to expand these and rewrite in terms of cot A and cot B.

    Try this and see how you get on.
 
 
 
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