# In ∆ABC,

Question:

In ∆ABC, if a2b2 and c2 are in A.P., prove that cot A, cot B and cot C are also in A.P.

Solution:

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

Then,

$\sin A=k a, \sin B=k b, \sin C=k c$

$a^{2}, b^{2}$ and $c^{2}$ are in A.P.

$\Rightarrow 2 b^{2}=a^{2}+c^{2}$

$\Rightarrow 2\left(a^{2}+c^{2}-b^{2}\right)=2\left(2 b^{2}-b^{2}\right)=2 b^{2}=b^{2}+b^{2}+c^{2}-a^{2}-c^{2}+a^{2}$

$\Rightarrow 2\left(a^{2}+c^{2}-b^{2}\right)=b^{2}+c^{2}-a^{2}+a^{2}+b^{2}-c^{2}$

$\Rightarrow \frac{2\left(a^{2}+c^{2}-b^{2}\right)}{2 a b c}=\frac{\left(b^{2}+c^{2}-a^{2}\right)}{2 a b c}+\frac{\left(a^{2}+b^{2}-c^{2}\right)}{2 a b c}$

$\Rightarrow \frac{2 \cos B}{k b}=\frac{\cos A}{k a}+\frac{\cos C}{k c}$

$\Rightarrow \frac{2 \cos B}{\sin B}=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}$

$\Rightarrow 2 \cot B=\cot A+\cot C$

$\Rightarrow \cot A, \cot B$ and $\cot C$ are in AP.