Question

In a $△ABC,a\cot A+b\cot B+c\cot C$ is equal to

• A. $r+R$
• B. $r-R$
• C. $2(r+R)$
• D. $2(r-R)$

Answer 1

The correct option is C $2(r+R)$

$a\cot A+b\cot B+c\cot C$
$=a\frac{\cos A}{\sin A}+b\frac{\cos B}{\sin B}+c\frac{\cos C}{\sin C}$
Using sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ we have
$=2R(\cos A+\cos B+\cos C)$
$=2R[2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}]$
$=2R[1+2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}]$
$=2R[1+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}]$
$=2R[1+\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right))]$
$=2R[1+\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2}))]$
$=2R[1+\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right))]$
Using transformation angle formula, we get
$=2R[1+\sin \left(\frac{C}{2}\right)2\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)]$
$=2R[1+4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)]$
$=2R(1+\frac{r}{R})$
$=2(R+r)$