Question

In a $△ABC,$ let $a,b$ and $c$ denote the lengths of sides opposite to vertices $A,B$ and $C$ respectively. If $a=7,b=3,c=5$ and
$\frac{ab}{(\tan \frac{B}{2}+\tan \frac{C}{2})(\tan \frac{A}{2}+\tan \frac{C}{2})}+\frac{bc}{(\tan \frac{A}{2}+\tan \frac{C}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})}+\frac{ac}{(\tan \frac{B}{2}+\tan \frac{C}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})}$
equals $k,$ then the value of $\sqrt{k}+\frac{1}{2}$ is

Answer 1

As we know,
$\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1$
$\frac{ab}{(\tan \frac{B}{2}+\tan \frac{C}{2})(\tan \frac{A}{2}+\tan \frac{C}{2})}=\frac{ab}{\tan \frac{A}{2}\tan \frac{C}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{B}{2}\tan \frac{A}{2}+{\tan }^2\frac{C}{2}}=\frac{ab}{1+{\tan }^2\frac{C}{2}}=s(s-c)(\because \cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}})$

Similarly,
$\frac{bc}{(\tan \frac{A}{2}+\tan \frac{C}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})}=s(s-a)$
and $\frac{ac}{(\tan \frac{B}{2}+\tan \frac{C}{2})(\tan \frac{A}{2}+\tan \frac{B}{2})}=s(s-b)$

Now, on adding
$s(s-c)+s(s-a)+s(s-b)=3s^2-s(a+b+c)=3s^2-2s^2=s^2=(7.5{)}^2$