what is the exact value of tan(195°)?\n\n$\frac{sqrt{3}+1}{1-sqrt{3}}$\n\n$\frac{sqrt{3}-1}{1+sqrt{3}}$\n\n$\…

Explanation:

Step1: Express (195^{\circ}) as a sum of two angles

We know that (195^{\circ}=150^{\circ}+45^{\circ}). The formula for (\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}). Here (A = 150^{\circ}) and (B=45^{\circ}). We know that (\tan150^{\circ}=\tan(180 - 30)^{\circ}=-\tan30^{\circ}=-\frac{\sqrt{3}}{3}) and (\tan45^{\circ}=1).

Step2: Substitute the values into the formula

[ \begin{align*} \tan(195^{\circ})&=\frac{\tan150^{\circ}+\tan45^{\circ}}{1-\tan150^{\circ}\tan45^{\circ}}\ &=\frac{-\frac{\sqrt{3}}{3}+ 1}{1-(-\frac{\sqrt{3}}{3})\times1}\ &=\frac{\frac{- \sqrt{3}+3}{3}}{\frac{3 + \sqrt{3}}{3}}\ &=\frac{3-\sqrt{3}}{3+\sqrt{3}}\ &=\frac{(3 - \sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3 - \sqrt{3})}\ &=\frac{9-6\sqrt{3}+3}{9-3}\ &=\frac{12-6\sqrt{3}}{6}\ & = 2-\sqrt{3} \end{align*} ] Now, rationalize (\frac{\sqrt{3}-1}{1 + \sqrt{3}}): [ \begin{align*} \frac{\sqrt{3}-1}{1+\sqrt{3}}&=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(1+\sqrt{3})(\sqrt{3}-1)}\ &=\frac{3-2\sqrt{3}+1}{3 - 1}\ &=\frac{4-2\sqrt{3}}{2}\ &=2-\sqrt{3} \end{align*} ]

Answer:

(\frac{\sqrt{3}-1}{1+\sqrt{3}}) (the second and third options in the given choices are same, and this is the correct one)