Given
[tex]\tan\frac{7\pi}{12}[/tex]
Rewrite as:
[tex]=\tan\left(\frac{\pi}{4}+\frac{\pi}{3}\right)[/tex]
Use the trigonometric identity for addition of angles:
[tex]=\frac{\tan\left(\frac{\pi}{4}\right)+\tan\left(\frac{\pi}{3}\right)}{1-\tan\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{3}\right)}[/tex]
Then:
[tex]\begin{gathered} \tan(\frac{\pi}{4})=1 \\ \tan(\frac{\pi}{3})=\sqrt{3} \end{gathered}[/tex]
Next, substitute the values:
[tex]=\frac{1+\sqrt{3}}{1-1\cdot\sqrt{3}}[/tex]
Simplify:
[tex]=\frac{1+\sqrt{3}}{1-\sqrt{3}}[/tex]
Multiply by numerator:
[tex]=\frac{\left(1+\sqrt{3}\right)\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}=\frac{1+\sqrt{3}+\sqrt{3}+3}{1+\sqrt{3}-\sqrt{3}-3}[/tex]
Simplify:
[tex]\frac{4+2\sqrt{3}}{-2}[/tex]
Factor the common term 2 in the numerator:
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Answer:
[tex]\begin{equation*} -2-\sqrt{3} \end{equation*}[/tex]
