In $ riangle A B C,(a+b+c)\left( an \frac{A}{2}+ an \frac{B}{2} ight)$ is equal to

Question

In $	riangle A B C,(a+b+c)\left(	an \frac{A}{2}+	an \frac{B}{2}ight)$ is equal to

Aicharya · Accepted Answer

We have, $(a+b+c)\left(	an \frac{A}{2}+	an \frac{B}{2}ight)$ $\begin{aligned} & =2 s\left(	an \frac{A}{2}+	an \frac{B}{2}ight) \ & =2\left(s 	an \frac{A}{2}+s 	an \frac{B}{2}ight) \ & =2\left(\frac{\Delta}{s-a}+\frac{\Delta}{s-b}ight) \ & =2 \Delta\left(\frac{s-b+s-a}{(s-a)(s-b)}ight) \ & =2 \Delta \frac{2 s-(a+b)}{(s-a)(s-b)} \ & =2 \Delta\left(\frac{c}{(s-a)(s-b)}ight) \ & =2 c \cot \frac{C}{2}\end{aligned}$

In $\triangle A B C,(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

Solution