Question

\(tan105^{\circ} =\) ......

• A. \(\frac{\sqrt{3}-1}{2\sqrt{2}}\)
• B. \(\frac{\sqrt{3}+1}{2\sqrt{2}}\)
• C. \(\frac{1+\sqrt{3}}{1-\sqrt{3}}\)
• D. \(\frac{2\sqrt{2}}{\sqrt{3}+1}\)

Hint:

Write 105 degrees as 60 plus 45 degrees and use the tangent addition formula.

Answer 1

The Correct Option is C

Step 1: Use the second quadrant.
$105^\circ = 90^\circ + 15^\circ$, so $\tan 105^\circ = -\cot 15^\circ$.

Step 2: Find cot 15.
$\tan 15^\circ = \tan(45^\circ - 30^\circ) = \dfrac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$. So $\cot 15^\circ = \dfrac{\sqrt{3}+1}{\sqrt{3}-1}$.

Step 3: Put the sign.
\[ \tan 105^\circ = -\frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{1+\sqrt{3}}{1-\sqrt{3}} \]

Final Answer:
Option (C). \[ \boxed{\frac{1+\sqrt{3}}{1-\sqrt{3}}} \]