Question

If A, B and C are angles in a triangle then\(\tan\left(\frac{A + B}{2}\right) \tan\left(\frac{C}{2}\right) \tan + \tan\left(\frac{B + C}{2}\right) \tan\left(\frac{A}{2}\right) + \tan\left(\frac{C + A}{2}\right) \tan\left(\frac{B}{2}\right)\) =

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is D

Let \( A + B + C = \pi \) (since these are angles in a triangle).
Then we know: \[ \frac{A+B}{2} = \frac{\pi - C}{2}, \quad \frac{B+C}{2} = \frac{\pi - A}{2}, \quad \frac{C+A}{2} = \frac{\pi - B}{2} \] Now using the identity: \[ \tan\left(\frac{\pi - x}{2}\right) = \cot\left(\frac{x}{2}\right) \] We get: \[ \tan\left(\frac{A+B}{2}\right) = \cot\left(\frac{C}{2}\right), \quad \tan\left(\frac{B+C}{2}\right) = \cot\left(\frac{A}{2}\right), \quad \tan\left(\frac{C+A}{2}\right) = \cot\left(\frac{B}{2}\right) \] Substitute into the expression: \[ \cot\left(\frac{C}{2}\right)\tan\left(\frac{C}{2}\right) + \cot\left(\frac{A}{2}\right)\tan\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right)\tan\left(\frac{B}{2}\right) \] Each term becomes 1: \[ 1 + 1 + 1 = 3 \]

The correct option is (D): \(3\)