Question

If \(0 < \alpha, \beta, \gamma < \frac{\pi}{2}\), such that \(\alpha + \beta + \gamma = \frac{\pi}{2}\) and \(\cot \alpha, \cot \beta, \cot \gamma\) are in AP, then the value of \(\cot \alpha \cot \gamma\) is

• A. 1
• B. 3
• C. \(\cot^2 \beta\)
• D. None of these

Hint:

Use \(\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}\) and \(\cot(\frac{\pi}{2} - \theta) = \tan \theta = \frac{1}{\cot \theta}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Given \(\cot \alpha, \cot \beta, \cot \gamma\) are in AP. Also \(\alpha + \beta + \gamma = \frac{\pi}{2}\).

Step 2: Detailed Explanation:
AP condition: \(2\cot \beta = \cot \alpha + \cot \gamma\).
Also \(\alpha + \gamma = \frac{\pi}{2} - \beta\).
Take cot: \(\cot(\alpha + \gamma) = \cot\left(\frac{\pi}{2} - \beta\right) = \tan \beta = \frac{1}{\cot \beta}\).
But \(\cot(\alpha + \gamma) = \frac{\cot \alpha \cot \gamma - 1}{\cot \alpha + \cot \gamma}\).
So \(\frac{\cot \alpha \cot \gamma - 1}{\cot \alpha + \cot \gamma} = \frac{1}{\cot \beta}\).
Let \(p = \cot \alpha + \cot \gamma = 2\cot \beta\) and \(q = \cot \alpha \cot \gamma\).
Then \(\frac{q - 1}{2\cot \beta} = \frac{1}{\cot \beta} \implies q - 1 = 2 \implies q = 3\).

Step 3: Final Answer:
\(\cot \alpha \cot \gamma = 3\). Option (B).