Find the value of \( \tan(105^\circ) \) using compound angle identities.
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Angles such as \(15^\circ, 75^\circ, 105^\circ\) are usually solved using compound angle identities like:
\[
\sin(A\pm B),\quad \cos(A\pm B),\quad \tan(A\pm B)
\]
Break the angle into known angles such as \(30^\circ, 45^\circ, 60^\circ\).
Topic: Trigonometry - Compound Angles Step 1: Understanding the Question:
We need to find the tangent value for an angle (\(105^\circ\)) that is not a standard value.
We must represent \(105^\circ\) as a sum of two standard angles. Step 2: Key Formula or Approach:
Use the identity:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
We can write \(105^\circ = 60^\circ + 45^\circ\). Step 3: Detailed Explanation:
1. Substitute \(A = 60^\circ\) and \(B = 45^\circ\):
\[ \tan(105^\circ) = \frac{\tan 60^\circ + \tan 45^\circ}{1 - \tan 60^\circ \tan 45^\circ} \]
2. Use standard values \(\tan 60^\circ = \sqrt{3}\) and \(\tan 45^\circ = 1\):
\[ \tan(105^\circ) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \]
3. Rationalize the denominator by multiplying by \((1 + \sqrt{3})\):
\[ \tan(105^\circ) = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \]
\[ = \frac{3 + 1 + 2\sqrt{3}}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} \]
\[ = -2 - \sqrt{3} = -(2 + \sqrt{3}) \] Step 4: Final Answer:
The value is \(-(2 + \sqrt{3})\).