Question

If circumradius and inradius of a triangle be $10$ and $3$ respectively, then the value of $a\cot A+b\cot B+c\cot C$ is equal to $25$

• A. True
• B. False

Answer 1

The correct option is B False

In $△ABC,$
$a\cot A+b\cot B+c\cot C$
$=2R\sin A\frac{\cos A}{\sin A}+2R\sin B\frac{\cos B}{\sin B}+2R\sin C\frac{\cos C}{\sin C}$
$=2R(\cos A+\cos B+\cos C)$
$=2R[2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}]$
$=2R[1+2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}]$
$=2R[1+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}]$
$=2R[1+2\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right))]$
$=2R[1+2\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2}))]$
$=2R[1+2\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right))]$
Using transformation angle formula, we get
$=2R[1+4\sin \left(\frac{C}{2}\right)2\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)]$
$=2R[1+4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)]$
$=2R(1+\frac{r}{R})$
$=2(R+r)$
$=2(R+r)$
$=2\times (10+3)=26$