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Math Help - tan, cot

  1. #1
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    tan, cot

    Hello,

    I would be happy if someone showed me how to solve this question

    a) By using the substitution t= tan 1/2 x
    prove that cosec x - cot x = tan 1/2 x

    b) Use this result to show that tan 15 degrees = 2 - sqrt(3)

    Thanks in advance.
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  2. #2
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    Hello Oasis1993
    Quote Originally Posted by Oasis1993 View Post
    Hello,

    I would be happy if someone showed me how to solve this question

    a) By using the substitution t= tan 1/2 x
    prove that cosec x - cot x = tan 1/2 x

    b) Use this result to show that tan 15 degrees = 2 - sqrt(3)

    Thanks in advance.
    The following formulae are standard results when \tan\tfrac12 x= t:
    \sin x = \frac{2t}{1+t^2}, \cos x=\frac{1-t^2}{1+t^2}, \tan x = \frac{2t}{1-t^2}
    So for (a):
    \csc x - \cot x = \frac{1}{\sin x} - \frac{1}{\tan x}
    =\frac{1+t^2}{2t}-\frac{1-t^2}{2t}
    I'm sure you can finish this now.

    (b) We know from a 30^o-60^o-90^o triangle, that \tan 30^o = \frac{1}{\sqrt3}. So, if t = \tan 15^o:
    \frac{2t}{1-t^2}=\frac{1}{\sqrt3}

    \Rightarrow 2\sqrt3t=1-t^2

    \Rightarrow t^2 +2\sqrt3t-1=0

    \Rightarrow t = \frac{-2\sqrt3 + \sqrt{(2\sqrt3)^2+4}}{2}, using the quadratic formula, and taking the positive result since \tan 15^o >0
    =\frac{-2\sqrt3+\sqrt{16}}{2}

    =2-\sqrt3
    Grandad
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