In a triangle \(\mathrm{ABC}\) with usual notations, if \(\tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C}\) are in H.P., then \(a^{2}, b^{2}, c^{2}\)
are in

If \(\tan A, \tan B, \tan C\) are in H.P., then
\(
\frac{2}{\tan B}=\frac{1}{\tan A}+\frac{1}{\tan C} \Rightarrow \frac{2 \cos B}{\sin B}=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}
\)
We know that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)
\(\therefore \frac{2\left(\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)}{b k}=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 b c}}{a k}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)
\(\Rightarrow \frac{2\left(c^{2}+a^{2}-b^{2}\right)}{2 a b c k}=\frac{b^{2}+c^{2}-a^{2}}{2 a b c k}+\frac{a^{2}+b^{2}-c^{2}}{2 a b c k}\)
\(2\left(a^{2}+c^{2}-b^{2}\right)=b^{2}+c^{2}-a^{2}+a^{2}+b^{2}-c^{2}\)
\(2 a^{2}+2 c^{2}-2 b^{2}=2 b^{2} \Rightarrow 4 b^{2}=2 a^{2}+2 c^{2} \Rightarrow 2 b^{2}\) \(=a^{2}+c^{2}\)
\(\Rightarrow a^{2}, b^{2}, c^{2} a r e\) in \(A \cdot P .\)