In a triangle $\mathrm{ABC}$ with usual notations, if $\tan \mathrm{A}, \tan \mathrm{B}, \tan \mathrm{C}$…
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
A.P
Not in A. P.
$\mathrm{H}. \mathrm{P}$
$\boldsymbol{G}.\mathrm{P}$
Solution
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 $A \cdot P .$