Question

If \( \alpha, \beta, \gamma \) are in AP and \( \tan^{-1}\alpha, \tan^{-1}\beta, \tan^{-1}\gamma \) are also in AP, then

• A. \( \alpha - \beta - \gamma = 0 \)
• B. \( \alpha = \beta = \gamma \)
• C. \( \alpha + \beta = \gamma \)
• D. \( 2\alpha = 3\beta = \gamma \)

Hint:

If both function and inverse function form AP $\longrightarrow$ equality is the safest conclusion.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
If three terms are in Arithmetic Progression (AP), twice the middle term equals the sum of the extreme terms. We apply this to both the values and their inverse tangents.
Step 2: Key Formula or Approach:
1. For \( \alpha, \beta, \gamma \) in AP: \( 2\beta = \alpha + \gamma \).
2. For \( \tan^{-1} \alpha, \tan^{-1} \beta, \tan^{-1} \gamma \) in AP: \( 2\tan^{-1} \beta = \tan^{-1} \alpha + \tan^{-1} \gamma \).
Step 3: Detailed Explanation:
From condition 2:
\[ \tan^{-1}\left(\frac{2\beta}{1 - \beta^2}\right) = \tan^{-1}\left(\frac{\alpha + \gamma}{1 - \alpha\gamma}\right) \]
Equating the arguments:
\[ \frac{2\beta}{1 - \beta^2} = \frac{\alpha + \gamma}{1 - \alpha\gamma} \]
Using condition 1 (\( 2\beta = \alpha + \gamma \)):
\[ \frac{2\beta}{1 - \beta^2} = \frac{2\beta}{1 - \alpha\gamma} \]
This gives two possibilities: \( \beta = 0 \) or \( 1 - \beta^2 = 1 - \alpha\gamma \).
If \( 1 - \beta^2 = 1 - \alpha\gamma \), then \( \beta^2 = \alpha\gamma \).
Since \( 2\beta = \alpha + \gamma \), \( \alpha, \beta, \gamma \) are in AP.
If \( \beta^2 = \alpha\gamma \), \( \alpha, \beta, \gamma \) are also in GP (Geometric Progression).
A sequence that is both AP and GP with non-zero terms must have all terms equal.
Thus, \( \alpha = \beta = \gamma \).
Step 4: Final Answer:
The relation is \(\alpha = \beta = \gamma\).