Solution
since
[tex]\begin{gathered} \sin(60)=\frac{\sqrt{3}}{2} \\ \\ \cos(60)=\frac{1}{2} \\ \\ \tan(60)=\sqrt{3} \\ \\ \sin(45)=\cos(45)=\frac{1}{\sqrt{2}} \\ \\ \tan(45)=1 \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow\sin(105)=\sin(60+45)=\sin(60)\cos(45)+\sin(45)\cos(60)=\frac{\sqrt{3}}{2}\times\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\times\frac{1}{2} \\ \\ =\frac{1}{2}\sqrt{\frac{3}{2}}+\frac{1}{2}\frac{1}{\sqrt{2}}=\frac{1}{2}(\frac{\sqrt{3}+1}{\sqrt{2}})=\frac{1}{2}(\frac{\sqrt{6}+\sqrt{2}}{2})=\frac{1}{4}(\sqrt{6}+\sqrt{2}) \end{gathered}[/tex]
[tex]\begin{gathered} \Rightarrow\cos(105)=\cos(60+45)=\cos(60)\cos(45)-\sin(60)\sin(45)=\frac{1}{2}\times\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\times\frac{1}{\sqrt{2}} \\ \\ =\frac{1}{2\sqrt{2}}-\frac{1}{2}\frac{\sqrt{3}}{\sqrt{2}}=\frac{1}{2}(\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}})=\frac{1}{2}(\frac{1-\sqrt{3}}{\sqrt{2}})=\frac{1}{4}(\sqrt{2}-\sqrt{6}) \end{gathered}[/tex][tex]\begin{gathered} \tan(105)=\frac{\sin(105)}{\cos(105)}=\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2}-\sqrt{6}}=\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2}-\sqrt{6}}\times\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2}+\sqrt{6}}=\frac{2+2\sqrt{12}+6}{2-6}=\frac{8+4\sqrt{3}}{-4} \\ \\ =-2-\sqrt{3} \end{gathered}[/tex]
