Question

If the sides $a,b,c$ of $△ABC$ are in $A.P.$, prove that
$\cos A\cot \frac{1}{2}A,\cos B\cot \frac{1}{2}B,\cos C\cot \frac{1}{2}C$ are in $A.P.$

Answer 1

We have $A\cot { \dfrac { 1 }{ 2 } } A=\left( 1-2\sin ^{ 2 }{ \dfrac { 1 }{ 2 } A } \right) .\dfrac { \cot { \dfrac { 1 }{ 2 } } A
}{ \cot { \dfrac { 1 }{ 2 } } A } $
$=\cot \frac{1}{2}A-\sin A$
Now $a,b,c$ are in $A.P.$
$\therefore \sin A,\sin B,\sin C$ are in also $A.P.$
Also we have proved in part $\left(i\right)$ that $\cot \left(A/2\right)$ are in $\cot \left(B/2\right)$, $\cot \left(C/2\right)$. Hence their differences are also in $A.P.$
Hence proved.