Question

If α+β+γ=2π, then the value of \(\cot\frac{\alpha}{2}\cot\frac{\beta}{2}+\cot\frac{\alpha}{2}\cot\frac{\gamma}{2}+\cot\frac{\beta}{2}\cot\frac{\gamma}{2}\) is

• A. 0
• B. 1
• C. \(\frac{\pi}{2}\)
• D. \(\frac{\pi}{3}\)
• E. \(\frac{1}{2}\)

Answer 1

The Correct Option is B

Step 1: Let \( A = \frac{\alpha}{2} \), \( B = \frac{\beta}{2} \), \( C = \frac{\gamma}{2} \). Then: \[ A + B + C = \pi \]

Step 2: Using the trigonometric identity for angles summing to \( \pi \): \[ \cot A \cot B + \cot A \cot C + \cot B \cot C = 1 \]

Step 3: Substitute back the original variables: \[ \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) + \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\gamma}{2}\right) + \cot\left(\frac{\beta}{2}\right)\cot\left(\frac{\gamma}{2}\right) = 1 \]

The correct answer is (B) 1.

Answer 2

Let the given expression be denoted as \( E \). 

We are given that \( \alpha + \beta + \gamma = 2\pi \). We want to find the value of:

\[ E = \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) + \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\gamma}{2}\right) + \cot\left(\frac{\beta}{2}\right)\cot\left(\frac{\gamma}{2}\right) \]

We can use the cotangent addition formula:

\[ \cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B} \]

Since \( \alpha + \beta + \gamma = 2\pi \), we have \( \alpha + \beta = 2\pi - \gamma \). Dividing by 2:

\[ \frac{\alpha + \beta}{2} = \pi - \frac{\gamma}{2} \]

Now, let's use the cotangent addition formula on \( \frac{\alpha}{2} + \frac{\beta}{2} \):

\[ \cot\left(\frac{\alpha + \beta}{2}\right) = \frac{\cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) - 1}{\cot\left(\frac{\alpha}{2}\right) + \cot\left(\frac{\beta}{2}\right)} \]

Substituting \( \frac{\alpha + \beta}{2} = \pi - \frac{\gamma}{2} \):

\[ \cot\left(\pi - \frac{\gamma}{2}\right) = -\cot\left(\frac{\gamma}{2}\right) = \frac{\cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) - 1}{\cot\left(\frac{\alpha}{2}\right) + \cot\left(\frac{\beta}{2}\right)} \]

Rearranging this equation:

\[ -\cot\left(\frac{\gamma}{2}\right)\left[\cot\left(\frac{\alpha}{2}\right) + \cot\left(\frac{\beta}{2}\right)\right] = \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) - 1 \] \[ -\cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\gamma}{2}\right) - \cot\left(\frac{\beta}{2}\right)\cot\left(\frac{\gamma}{2}\right) = \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) - 1 \]

Adding \( \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) \) to both sides and rearranging:

\[ \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\beta}{2}\right) + \cot\left(\frac{\alpha}{2}\right)\cot\left(\frac{\gamma}{2}\right) + \cot\left(\frac{\beta}{2}\right)\cot\left(\frac{\gamma}{2}\right) = 1 \]

Therefore, the value of the given expression is 1.