Question

Prove that $\frac{\tan \frac{1}{2}A}{(a-b)(a-c)}+\frac{\tan \frac{1}{2}B}{(b-c)(b-a)}+\frac{\tan \frac{1}{2}C}{(c-a)(c-b)}=\frac{1}{S}$.

Answer 1

Putting $\tan \frac{1}{2}A=\sqrt{\left(\frac{(s-b)(s-c)}{s(s-a)}\right)}$ etc.,
We have
$=[\frac{\sqrt{\left(\frac{(s-b)(s-c)}{s(s-a)}\right)}}{(a-b)(a-c)}+\frac{\sqrt{\left(\frac{(s-c)(s-a)}{s(s-b)}\right)}}{(b-c)(b-a)}+\frac{\sqrt{\left(\frac{(s-a)(s-b)}{s(s-c)}\right)}}{(c-a)(c-b)}]$
$=\frac{(c-b)(s-b)(s-c)+(a-c)(s-c)(s-a)+(b-a)(s-a)(s-c)}{(a-b)(b-c)(c-a)\surd \left[s(s-a)(s-b)(s-c)\right]}$
$N^r=-\sum \{(b-c)\{s^2-s(b+c)+bc\}$
$=-s^2\sum (b-c)+s\sum (b^2-c^2)-\sum bc(b-c)$
$=0+0(b-c)(c-a)(a-b)$
Note that on simplification,
$bc(b-c)+ca(c-a)+ab(a-b)$
$=-(b-c)(c-a)(a-b)]$
Substituting in $\left(1\right)$, we gat
$L.H.S=\frac{N^r}{D^r}=\frac{1}{\surd \left[s\right(s-a\left)\right(s-b\left)\right(s-c\left)\right]}=\frac{1}{S}$