Use The Half-Angle Formulas
-$\tan 165 ^ { \circ }$
A) $\sqrt { 3 } - 2$
B) $\sqrt { 3 } + 2$
C) $- \sqrt { 3 } + 2$
D) $- \sqrt { 3 } - 2$

Question

Quizplus · Accepted Answer

The half-angle formula for tangent is $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$. Using this formula, we have $\tan 82.5 ^ { \circ } = \frac{\sin 165 ^ { \circ }}{1 + \cos 165 ^ { \circ }}$. Since $\sin 165 ^ { \circ } = \sin (180 ^ { \circ } - 15 ^ { \circ }) = \sin 15 ^ { \circ }$ and $\cos 165 ^ { \circ } = -\cos 15 ^ { \circ }$, we have $\tan 82.5 ^ { \circ } = \frac{\sin 15 ^ { \circ }}{1 - \cos 15 ^ { \circ }}$. Using the half-angle formula for cosine, we have $\cos 15 ^ { \circ } = \sqrt{\frac{1 + \cos 30 ^ { \circ }}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$. Substituting this into the expression for $\tan 82.5 ^ { \circ }$, we have \(\tan 82.5 ^ { \circ } = \frac{\sin 15 ^ { \circ }}{1 - \frac{\sqrt{2 + \sqrt{3}}}{2}} = \frac{\frac{1 - \sqrt{3}}{2}}{1 - \frac{\sqrt{2 + \sqrt{3}}}{2}} = \frac{1 - \sqrt{3}}{2 - \sqrt{2 + \sqrt{3}}} = \frac{(1 - \sqrt{3})(2 + \sqrt{2 + \sqrt{3}})}{(2 - \sqrt{2 + \sqrt{3}})(2 + \sqrt{2 + \sqrt{3}})} = \frac{2 + \sqrt{2 + \sqrt{3}} - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}}}{4 - 2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}} - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}}}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 - 2\sqrt{3} - \sqrt{6 + 3\sqrt{3}})}{2 + \sqrt{3}} = \frac{\sqrt{2 + \sqrt{3}}(1 -

Use the Half-Angle Formulas - tan⁡165∘\tan 165 ^ { \circ }tan165∘

Use the Half-Angle Formulas
- $\tan 165 ^ { \circ }$