Remember that
[tex]tan(\frac{7\pi}{12})=tan(105^o)=-tan(75^o)=-tan(45^o+30^o)[/tex]
and
[tex]tan(A+B)=\frac{tanA+tanB}{1-tanA*tanB}[/tex]
therefore
[tex]tan(45^o+30^o)=\frac{tan45^o+tan30^o}{1-tan45^o*tan30^o}[/tex]
substitute given values
[tex]tan(45^o+30^o)=\frac{1+\frac{\sqrt{3}}{3}}{1-(1)(\frac{\sqrt{3}}{3})}=\frac{\frac{3+\sqrt{3}}{3}}{\frac{3-\sqrt{3}}{3}}=\frac{3+\sqrt{3}}{3-\sqrt{3}}*\frac{3+\sqrt{3}}{3+\sqrt{3}}=\frac{9+6\sqrt{3}+3}{6}=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}[/tex]
therefore
The answer is
[tex]tan(\frac{7\pi}{12})=-(2+\sqrt{3})[/tex]
The answer is the second option
