Question

If $R,r$ denote respectively the radius of the circumcircle and incircle of a triangle ABC, then

Answer 1

From laws of sin
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$ ...(1)
In $△ABC$ $A+B+C=\pi$ ...(2)
Using (1) and (2)
$a\cos A+b\cos B+c\cos C=2R(\sin A\cos A+\sin B\cos B+\sin C\cos C)=R(\sin 2A+\sin 2B+\sin 2C)=R(2\sin (A+B)\cos (A-B)+\sin (2\pi -2(A+B)))=R(2\sin (A+B)\cos (A-B)-2\sin (A+B)\cos (A+B))=2R\sin (A+B)(\cos (A-B)-\cos (A+B))=2R\sin (\pi -C)\left(2\sin A\sin B\right)=4R\sin A\sin B\sin C$
Now,
$a\cot A+b\cot B+c\cot C=2R(\cos A+\cos B+\cos C)=2R(1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})$ ...(3)
As $△=sr$ ...(4)
And as $△=\frac{1}{2}ab\sin C$
From (1)
$△=\frac{1}{2}2R\sin A2R\sin B\sin C=2R^2\sin A\sin B\sin C$ ...(5)
Also $s=(a+b+c)=R(\sin A+\sin B+\sin C)=4R\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$ ...(6)
From (4), (5) and (6)
$4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=r$
Substituting this in (3)
$a\cot A+b\cot B+c\cot C=2R+2r$
And now from (1)
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
Gives
$\frac{a}{\sin A}+\frac{b}{\sin B}+\frac{c}{\sin C}=6R$