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Math Help - A Sum and difference question

  1. #1
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    A Sum and difference question

    Tan(\frac{4\pi}{3}-\frac{\pi}{4}) = \frac{Tan(\frac{4\pi}{3})-Tan(\frac{\pi}{4})}{1+Tan(\frac{4\pi}{3}Tan(\frac{  \pi}{4})}=\frac{\sqrt{3}-1}{1+\sqrt{3}*1}= I do not know this step = 2-\sqrt{3}

    Could someone tell me what happens to take it from \frac{\sqrt{3}-1}{1+\sqrt{3}*1} = 2-\sqrt{3}
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  2. #2
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    Quote Originally Posted by Dannbr View Post
    Tan(\frac{4\pi}{3}-\frac{\pi}{4}) = \frac{Tan(\frac{4\pi}{3})-Tan(\frac{\pi}{4})}{1+Tan(\frac{4\pi}{3}Tan(\frac{  \pi}{4})}=\frac{\sqrt{3}-1}{1+\sqrt{3}*1}= I do not know this step = 2-\sqrt{3}

    Could someone tell me what happens to take it from \frac{\sqrt{3}-1}{1+\sqrt{3}*1} = 2-\sqrt{3}
    Multiply numerator and denominator by conjugate of denominator.

    \displaystyle \frac{\sqrt{3}-1}{1+\sqrt{3}}=\frac{(\sqrt{3}-1)(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\cdots=2-\sqrt{3}
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