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: Understanding the Concept:
We can find \(\tan 105^\circ\) by splitting the angle into two angles whose tangent is known.

Step 2: Key Formula or Approach:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

Step 3: Detailed Explanation:
Take \(A = 60^\circ\) and \(B = 45^\circ\), with \(\tan 60^\circ = \sqrt{3}\) and \(\tan 45^\circ = 1\).
\[ \tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}\cdot 1} = \frac{1+\sqrt{3}}{1-\sqrt{3}} \]
This matches option (C). The value is negative, as expected for an angle in the second quadrant.

Step 4: Why the other options are wrong.
(A) and (B) are both positive numbers, but \(\tan 105^\circ\) is negative. (D) is also positive.

Final Answer:
\(\tan 105^\circ = \dfrac{1+\sqrt{3}}{1-\sqrt{3}}\), option (C). \[ \boxed{\frac{1+\sqrt{3}}{1-\sqrt{3}}} \]