Question

In ∆ABC, if a², b² and c² are in A.P., prove that cot A, cot B and cot C are also in A.P.

Answer 1

$\mathrm{Let}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

Then,
$\sin A=ka,\sin B=kb,\sin C=kc$
a², b² and c² are in A.P.
$$\begin{gathered} \Rightarrow 2b^2=a^2+c^2 \\ \Rightarrow 2\left(a^2+c^2-b^2\right)=2\left(2b^2-b^2\right)=2b^2=b^2+b^2+c^2-a^2-c^2+a^2 \\ \Rightarrow 2\left(a^2+c^2-b^2\right)=b^2+c^2-a^2+a^2+b^2-c^2 \\ \Rightarrow \frac{2\left(a^2+c^2-b^2\right)}{2abc}=\frac{\left(b^2+c^2-a^2\right)}{2abc}+\frac{\left(a^2+b^2-c^2\right)}{2abc} \\ \Rightarrow \frac{2\cos B}{kb}=\frac{\cos A}{ka}+\frac{\cos C}{kc} \\ \Rightarrow \frac{2\cos B}{\sin B}=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C} \\ \Rightarrow 2\cot B=\cot A+\cot C \\ \Rightarrow \cot A,\cot B\mathrm{and}\cot C\mathrm{are}\mathrm{in}\mathrm{AP}. \end{gathered}$$