Question

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

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

Answer 1

The correct option is A

$\frac{7}{9}$

Finding $\tan \left(\frac{C}{2}\right)$:

Let $A$, $B$ and $C$ are the angles of a plain triangle

Sum of the angles of the triangle is $180°$

Thus,$A+B+C=\mathrm{\pi }$

Dividing by $2$ on both sides

$$\begin{gathered} \frac{A+B+C}{2}=\frac{\mathrm{\pi }}{2} \\ \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\mathrm{\pi }}{2} \\ \frac{A}{2}+\frac{B}{2}=\frac{\mathrm{\pi }}{2}-\frac{C}{2} \end{gathered}$$

Taking $\tan$ on both sides we get

$$\begin{gathered} \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\mathrm{\pi }}{2}-\frac{C}{2}\right) \\ \frac{\tan \left({\displaystyle \frac{A}{2}}\right)+\tan \left({\displaystyle \frac{B}{2}}\right)}{1-\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)}=\cot \left(\frac{C}{2}\right)\left[\because \tan \left(A+B\right)=\frac{\tan \left({\displaystyle A}\right)+\tan \left({\displaystyle B}\right)}{1-\tan \left(A\right)\tan \left(B\right)};\tan \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\cot \left(\theta \right)\right] \end{gathered}$$

$$\begin{gathered} \frac{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{2}{3}}}{1-\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{2}{3}}\right)}=\cot \left(\frac{C}{2}\right) \\ \frac{{\displaystyle 1}}{1-{\displaystyle \frac{2}{9}}}=\cot \left(\frac{C}{2}\right) \\ \frac{{\displaystyle 9}}{9-2}=\cot \left(\frac{C}{2}\right) \\ \frac{9}{7}=\frac{1}{\tan \left(\frac{C}{2}\right)}\left[\because \cot \left(x\right)=\frac{1}{\tan \left(x\right)}\right] \\ \tan \left(\frac{C}{2}\right)=\frac{7}{9} \end{gathered}$$

Hence, option (A) is the correct answer.