Recall the rules
[tex]\sin(a-b)=\sin(a) \cos(b) - \cos(a) \sin(b)[/tex]
[tex]\cos(a-b)=\sin(a) \sin(b) + \cos(a) \cos(b)[/tex]
Use the suggested difference:
[tex]\sin(195)=\sin(225-30)=\sin(225) \cos(30) - \cos(225) \sin(30)[/tex]
[tex]\cos(195)=\cos(225-30)=\sin(225) \sin(30) + \cos(225) \cos(30)[/tex]
Since 225 and 30 are known angles, we can plug the values:
[tex]\sin(225)=\cos(225)=-\dfrac{1}{\sqrt{2}},\quad \sin(30)=\dfrac{1}{2},\quad \cos(30)=\dfrac{\sqrt{3}}{2}[/tex]
The expressions become
[tex]\sin(195)=-\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} - \left(-\dfrac{1}{\sqrt{2}}\right) \cdot \dfrac{1}{2} = \dfrac{1-\sqrt{3}}{2\sqrt{2}}[/tex]
[tex]\cos(195)=-\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} + \left(-\dfrac{1}{\sqrt{2}}\right) \dfrac{\sqrt{3}}{2}=\dfrac{-1-\sqrt{3}}{2\sqrt{2}}[/tex]
As usual, we just use the definition of the tangent:
[tex]\tan(195)=\dfrac{\sin(195)}{\cos(195)}=\dfrac{\frac{1-\sqrt{3}}{2\sqrt{2}}}{\frac{-1-\sqrt{3}}{2\sqrt{2}}}=\dfrac{1-\sqrt{3}}{-1-\sqrt{3}}[/tex]
