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Math Help - Tan - equation

  1. #1
    Super Member dhiab's Avatar
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    Tan - equation

    Prove :
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  2. #2
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    \tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(\frac{5\pi}{12}\right)
    \begin{aligned}<br />
&= \tan^2\left(\frac{\pi}{12}\right) + \cot^2\left(\frac{\pi}{12}\right) \\<br />
&= \left[\csc \left(\frac{\pi}{6}\right) - \cot \left(\frac{\pi}{6}\right)\right]^2 + \left[\csc \left(\frac{\pi}{6}\right) + \cot \left(\frac{\pi}{6}\right)\right]^2<br />
\end{aligned}
    \begin{aligned}<br />
&= \csc^2 \left(\frac{\pi}{6}\right) - 2\csc \left(\frac{\pi}{6}\right)\cot \left(\frac{\pi}{6}\right) + \cot^2 \left(\frac{\pi}{6}\right) \\<br />
&\;\;+ \csc^2 \left(\frac{\pi}{6}\right) + 2\csc \left(\frac{\pi}{6}\right)\cot \left(\frac{\pi}{6}\right) + \cot^2 \left(\frac{\pi}{6}\right) \\<br />
&= 2\csc^2 \left(\frac{\pi}{6}\right) + 2\cot^2 \left(\frac{\pi}{6}\right) \\<br />
\end{aligned}
    \begin{aligned}<br />
&= 2\csc^2 \left(\frac{\pi}{6}\right) + 2\left(\csc^2 \left(\frac{\pi}{6}\right) - 1\right) \\<br />
&= 4\csc^2 \left(\frac{\pi}{6}\right) - 2 \\<br />
&= 4(2^2) - 2 \\<br />
&= 14<br />
\end{aligned}

    Note: this was not meant to be an efficient way.


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  3. #3
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    Hello, dhiab!

    My method is not very efficient either . . .

    . . \begin{array}{ccc}\sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} \\ \\[-4mm] \cos^2\!\theta &=& \dfrac{1+\cos2\theta}{2} \end{array}



    \text{Prove: }\;\tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(\frac{5\pi}{12}\right) \:=\:14

    We have: . \frac{\sin^2(\frac{\pi}{12})}{\cos^2(\frac{\pi}{12  })} + \frac{\sin^2(\frac{5\pi}{12})}{\cos^2(\frac{5\pi}{  12})} . =\;\; \frac{\dfrac{1-\cos\frac{\pi}{6}}{2}} {\dfrac{1 + \cos\frac{\pi}{6}}{2}}  +  \frac{\dfrac{1-\cos\frac{5\pi}{6}}{2}} {\dfrac{1 + \cos\frac{5\pi}{6}}{2}}

    . . . . . . = \;\frac{1-\cos\frac{\pi}{6}}{1 + \cos\frac{\pi}{6}} + \frac{1-\cos\frac{5\pi}{6}}{1 + \cos\frac{5\pi}{6}} \;=\;\frac{1-\frac{\sqrt{3}}{2}} {1 + \frac{\sqrt{3}}{2}} + \frac{1+\frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}

    . . . . . . =\; \frac{\left(1-\frac{\sqrt{3}}{2}\right)^2 + \left(1 + \frac{\sqrt{3}}{2}\right)^2} {1 - \left(\frac{\sqrt{3}}{2}\right)^2} \;=\; \frac{\left(1 - \sqrt{3} + \frac{3}{4}\right) + \left(1 + \sqrt{3} + \frac{3}{4}\right)}{1 - \frac{3}{4}}

    . . . . . . =\;\frac{\frac{14}{4}}{\frac{1}{4}} \;=\;\boxed{14}

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