Question

In ΔABC, A + B + C = π show that

${\displaystyle \tan \frac{\text{A}}{2}\tan \frac{\text{B}}{2}+\tan \frac{\text{B}}{2}\tan \frac{\text{C}}{2}+\tan \frac{\text{C}}{2}\tan \frac{\text{A}}{2}}$ = 1

Answer 1

In ΔABC, A + B + C = π

∴ ${\displaystyle \frac{\text{A}}{2}+\frac{\text{B}}{2}+\frac{\text{C}}{2}=\frac{\pi }{2}}$

∴ ${\displaystyle \frac{\text{A}}{2}+\frac{\text{B}}{2}=\frac{\pi }{2}-\frac{\text{C}}{2}}$

∴ ${\displaystyle \tan (\frac{\text{A}}{2}+\frac{\text{B}}{2})=\tan (\frac{\pi }{2}-\frac{\text{C}}{2})}$

∴ ${\displaystyle \frac{\tan \left(\frac{\text{A}}{2}\right)+\tan \left(\frac{\text{B}}{2}\right)}{1-\tan \left(\frac{\text{A}}{2}\right)\cdot \tan \left(\frac{\text{B}}{2}\right)}=\cot \left(\frac{\text{C}}{2}\right)=\frac{1}{\tan \left(\frac{\text{C}}{2}\right)}}$

∴ ${\displaystyle \tan \left(\frac{\text{C}}{2}\right)\cdot \tan \left(\frac{\text{A}}{2}\right)+\tan \left(\frac{\text{B}}{2}\right)\cdot \tan \left(\frac{\text{C}}{2}\right)=1-\tan \left(\frac{\text{A}}{2}\right)\cdot \tan \left(\frac{\text{B}}{2}\right)}$

∴ ${\displaystyle \tan \left(\frac{\text{A}}{2}\right)\cdot \tan \left(\frac{\text{B}}{2}\right)+\tan \left(\frac{\text{B}}{2}\right)\cdot \tan \left(\frac{\text{C}}{2}\right)+\tan \left(\frac{\text{C}}{2}\right)\cdot \tan \left(\frac{\text{A}}{2}\right)}$ = 1.