Find the exact value by using a half-angle identity.
-$\tan 345 ^ { \circ }$
A) $\sqrt { 3 } - 2$
B) $- \sqrt { 3 } + 2$
C) $\sqrt { 3 } + 2$
D) $- \sqrt { 3 } - 2$

Question

Quizplus · Accepted Answer

The half-angle identity for tangent is $\tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$. Since $345^\circ$ is in the fourth quadrant, $\tan 345^\circ$ is negative. So, we have$\tan 345^\circ = - \sqrt{\frac{1 - \cos 345^\circ}{1 + \cos 345^\circ}}$$\cos 345^\circ = \cos (360^\circ - 15^\circ) = \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$So,$\tan 345^\circ = - \sqrt{\frac{1 - \frac{\sqrt{6} + \sqrt{2}}{4}}{1 + \frac{\sqrt{6} + \sqrt{2}}{4}}}$$\tan 345^\circ = - \sqrt{\frac{4 - \sqrt{6} - \sqrt{2}}{4 + \sqrt{6} + \sqrt{2}}}$$\tan 345^\circ = - \sqrt{\frac{(2 - \sqrt{3})(2 + \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}}$$\tan 345^\circ = - \sqrt{2 - \sqrt{3}}$$\tan 345^\circ = - (\sqrt{3} - 2)$$\tan 345^\circ = - \sqrt{3} + 2$

Find the Exact Value by Using a Half-Angle Identity tan⁡345∘\tan 345 ^ { \circ }tan345∘

Find the Exact Value by Using a Half-Angle Identity $\tan 345 ^ { \circ }$