Question

If $A+B+C=\pi$, prove that $\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1$.

Answer 1

It is given that $A+B+C=\pi$, then,

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

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

$\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi }{2}-\frac{C}{2}\right)$

$\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}=\cot \frac{C}{2}$

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

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

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