Question

In any $△ABC$, if $a^2,b^2,c^2$ are in $A.P$., then prove that $\cot A,\cot B,\cot C$ are in $A.P$.

Answer 1

Given:$a^2,b^2,c^2$ are in A.P

$\Rightarrow b^2-a^2=c^2-b^2$

$\Rightarrow k^2{\sin }^{2}B-k^2{\sin }^{2}A=k^2{\sin }^{2}C-k^2{\sin }^{2}B$ using sine rule $a=k\sin A,b=k\sin B$ and $c=k\sin C$

$\Rightarrow \sin (B+A)\sin (B-A)=\sin (C+B)\sin (C-B)$

$\Rightarrow \sin (\pi -C)\sin (B-A)=\sin (\pi -A)\sin (C-B)$ since $A+B+C=\pi$

$\Rightarrow \sin C\sin (B-A)=\sin A\sin (C-B)$

$\Rightarrow \frac{\sin (B-A)}{\sin A}=\frac{\sin (C-B)}{\sin C}$

$\Rightarrow \frac{\sin (B-A)}{\sin A\sin B}=\frac{\sin (C-B)}{\sin B\sin C}$

$\Rightarrow \frac{\sin B\cos A-\cos B\sin A}{\sin A\sin B}=\frac{\sin C\cos B-\cos B\sin C}{\sin B\sin C}$

$\Rightarrow \cot A-\cot B=\cot B-\cot C$

$\therefore \cot A,\cot B,\cot C$ are in A.P

hence Proved