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Thread: 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 $\displaystyle \tan\tfrac12 x= t$:
    $\displaystyle \sin x = \frac{2t}{1+t^2}, \cos x=\frac{1-t^2}{1+t^2}, \tan x = \frac{2t}{1-t^2}$
    So for (a):
    $\displaystyle \csc x - \cot x = \frac{1}{\sin x} - \frac{1}{\tan x}$
    $\displaystyle =\frac{1+t^2}{2t}-\frac{1-t^2}{2t}$
    I'm sure you can finish this now.

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

    $\displaystyle \Rightarrow 2\sqrt3t=1-t^2$

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

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

    $\displaystyle =2-\sqrt3$
    Grandad
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