Question

In \( \triangle ABC \), \( (\cot A + \cot B)(\cot B + \cot C)(\cot C + \cot A) = \)

• A. \( \sec A \sec B \sec C \)
• B. \( \tan A \tan B \tan C \)
• C. \( \mathbf{\csc A \csc B \csc C} \)
• D. \( \cot A \cot B \cot C \)

Hint:

Use standard identities and sum of angles in triangle to simplify expressions involving cotangents.

Answer 1

The Correct Option is C

In any triangle \( \triangle ABC \), it holds that: \[ \cot A + \cot B = \frac{\sin(A + B)}{\sin A \sin B} = \frac{\sin C}{\sin A \sin B} \] Similarly, \[ (\cot A + \cot B)(\cot B + \cot C)(\cot C + \cot A) = \frac{\sin C}{\sin A \sin B} \cdot \frac{\sin A}{\sin B \sin C} \cdot \frac{\sin B}{\sin A \sin C} \] Multiplying: \[ = \frac{\sin C \cdot \sin A \cdot \sin B}{(\sin A \sin B)^2 \cdot \sin C^2} = \frac{1}{\sin A \sin B \sin C} = \csc A \csc B \csc C \]