Question

If $\alpha+\beta+\gamma=2\pi$, then $\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\gamma}{2} = $ ________.

• A. $\tan\frac{\alpha}{2} \tan\frac{\beta}{2} \tan\frac{\gamma}{2}$
• B. $\tan\frac{\alpha}{2} \tan\frac{\beta}{2} \tan\frac{\gamma}{2}$ (Duplicate)
• C. $2\tan\frac{\alpha}{2} \tan\frac{\beta}{2} \tan\frac{\gamma}{2}$
• D. $3\tan\frac{\alpha}{2} \tan\frac{\beta}{2} \tan\frac{\gamma}{2}$
• E. $4\tan\frac{\alpha}{2} \tan\frac{\beta}{2} \tan\frac{\gamma}{2}$

Hint:

Sum = Product for tangents of angles that sum to $\pi$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
This problem uses a standard trigonometric identity related to the sum of tangents of three angles that add up to \(\pi\). We first need to transform the given condition to match the form required by the identity.
Step 2: Key Formula or Approach
The key identity is: If \(A+B+C = \pi\), then \(\tan A + \tan B + \tan C = \tan A \tan B \tan C\).
We are given \(\alpha + \beta + \nu = 2\pi\). We will manipulate this equation to fit the form \(A+B+C=\pi\).
Step 3: Detailed Explanation
1. Transform the given condition.
We are given:
\[ \alpha + \beta + \nu = 2\pi \] Divide the entire equation by 2:
\[ \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\nu}{2} = \pi \] 2. Apply the trigonometric identity.
Let \(A = \frac{\alpha}{2}\), \(B = \frac{\beta}{2}\), and \(C = \frac{\nu}{2}\).
Our condition now becomes \(A+B+C = \pi\).
According to the identity, if three angles sum to \(\pi\), the sum of their tangents is equal to the product of their tangents.
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \] 3. Substitute the original angles back.
Replacing A, B, and C with their expressions in terms of \(\alpha, \beta, \nu\):
\[ \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\nu}{2} = \tan\frac{\alpha}{2} \tan\frac{\beta}{2} \tan\frac{\nu}{2} \] Derivation of the identity (for completeness):
From \(A+B+C = \pi\), we have \(A+B = \pi - C\).
Take the tangent of both sides:
\[ \tan(A+B) = \tan(\pi - C) \] Using the tangent addition formula and the property \(\tan(\pi - \theta) = -\tan\theta\):
\[ \frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C \] Multiply both sides by \((1 - \tan A \tan B)\):
\[ \tan A + \tan B = -\tan C (1 - \tan A \tan B) \] \[ \tan A + \tan B = -\tan C + \tan A \tan B \tan C \] Rearrange the terms to get the identity:
\[ \tan A + \tan B + \tan C = \tan A \tan B \tan C \] Step 4: Final Answer
The expression is equal to \(\tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\nu}{2}\).