use a half - angle formula to evaluate the expression without a calculator. tan(-105°) (type an exact…

Explanation:

Step1: Recall the half - angle formula for tangent

The half - angle formula for tangent is $\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1 - \cos\alpha}{1+\cos\alpha}}$. First, note that $\tan(-105^{\circ})=-\tan(105^{\circ})$. And $105^{\circ}=\frac{210^{\circ}}{2}$.

Step2: Determine the sign

Since $105^{\circ}$ is in the second quadrant where tangent is negative, for $\tan105^{\circ}=\tan\frac{210^{\circ}}{2}$, we take the negative sign in the half - angle formula. So $\tan105^{\circ}=-\sqrt{\frac{1-\cos210^{\circ}}{1 + \cos210^{\circ}}}$, and $\tan(-105^{\circ})=\sqrt{\frac{1-\cos210^{\circ}}{1+\cos210^{\circ}}}$. We know that $\cos210^{\circ}=-\frac{\sqrt{3}}{2}$.

Step3: Substitute the value of cosine

Substitute $\cos210^{\circ}=-\frac{\sqrt{3}}{2}$ into the formula: [ \begin{align*} \tan(-105^{\circ})&=\sqrt{\frac{1-(-\frac{\sqrt{3}}{2})}{1+(-\frac{\sqrt{3}}{2})}}\ &=\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}\ &=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}}}\ &=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \end{align*} ] Rationalize the denominator: [ \begin{align*} \tan(-105^{\circ})&=\sqrt{\frac{(2 + \sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\ &=\sqrt{\frac{4 + 4\sqrt{3}+3}{4-3}}\ &=2+\sqrt{3} \end{align*} ] In terms of the half - angle formula form, $\tan(-105^{\circ})=\sqrt{\frac{1-\cos210^{\circ}}{1+\cos210^{\circ}}}$, so the correct option is C.

Answer:

C. $\tan(-105^{\circ})=-\sqrt{\frac{1-\cos210^{\circ}}{1+\cos210^{\circ}}}$