Question

Let \(A\), \(B\) and \(C\) are the angles of a triangle and \(\tan \frac{A}{2} = 1/3\), \(\tan \frac{B}{2} = \frac{2}{3}\). Then, \(\tan \frac{C}{2}\) is equal to:

• A. \(\frac{7}{9}\)
• B. \(\frac{2}{9}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{2}{3}\)

Hint:

Always verify angle sum properties and recalculations when dealing with trigonometric identities in geometry.

Answer 1

The Correct Option is A

Using the angle sum property of a triangle: \[ A + B + C = 180^\circ \] \[ \Rightarrow C = 180^\circ - A - B \] Given: \[A + B + C = 180^\circ \quad {(Angle sum property of a triangle)}\] \[\Rightarrow A + B = 180^\circ - C\] \[\Rightarrow \frac{A+B}{2} = 90^\circ - \frac{C}{2}\] \[\Rightarrow \tan\left(\frac{A}{2} + \frac{B}{2}\right) = \tan\left(90^\circ - \frac{C}{2}\right)\] \[\Rightarrow \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \cot \frac{C}{2}\] \[\Rightarrow \frac{\frac{1}{3} + \frac{2}{3}}{1 - \frac{1}{3} \times \frac{2}{3}} = \cot \frac{C}{2}\] \[\Rightarrow \frac{1}{1 - \frac{2}{9}} = \cot \frac{C}{2}\] \[ \Rightarrow \frac{9}{7} = \cot \frac{C}{2} \]\[\Rightarrow \tan \frac{C}{2} = \frac{7}{9}\]