Question

$a\cot A+b\cot B+c\cot C=2(R+r)$

Answer 1

LHS: $a\cot A+b\cot B+c\cot C$

$=2R\sin A\frac{\cos A}{\sin A}+2R\sin B\frac{\cos B}{\sin B}+2R\sin C\frac{\cos C}{\sin C}$ ------Since, $a=2R\sin A$

$=2R(\cos A+\cos B+\cos C)$---(i)

Consider, $\cos A+\cos B+\cos C$

$=(\cos A+\cos B)+\cos C$

$=2\cos \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\left(\frac{C}{2}\right)$

$=2\cos \left(\frac{180-C}{2}\right).\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\left(\frac{C}{2}\right)+1$

$=2\sin \left(\frac{C}{2}\right).\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\left(\frac{C}{2}\right)+1$

$=2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right)]+1$

$=2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{180-(A+B)}{2}\right)]+1$

$=2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]+1$

$=2\sin \left(\frac{C}{2}\right).2\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)+1$

$=\frac{4R\sin \left(\frac{C}{2}\right).\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)}{R}+1$

$=\frac{r}{R}+1$

Substitute in (i)

$=2R(\frac{r}{R}+1)$

$=2R\left(\frac{r+R}{R}\right)$

$=2(r+R)$ :RHS