Proving $\cot { A+\cot { B+\cot { C=\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 4K } } } } $

Question

For any acute $\triangle ABC$, prove that $\cot { A+\cot { B+\cot { C=\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 4K } } } } $, where $K$ is the area of $\triangle ABC$.

Unfortunately I'm not able to progress in this problem. Any kind help will be appreciated.

Thank you.

Answer 1

Here is a straightforward solution

$\cot A+\cot B+\cot C = \frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}=\frac{(b^2+c^2-a^2)}{2bc\frac{a}{2R}}+\frac{(a^2+c^2-b^2)}{2ac\frac{b}{2R}}+\frac{(a^2+b^2-c^2)}{2ab\frac{c}{2R}}$.

Now use $\frac{abc}{4R}=K$

Answer 2

$$\cot A=\dfrac{\cos A}{\sin A}=\dfrac{b^2+c^2-a^2}{2abc}$$

Now $\triangle =\dfrac{abc}{4R}$

Answer 3

Hint: $4K = 4\cdot \dfrac{abc}{4R}= \dfrac{abc}{R}= 2bc\sin A\Rightarrow 4K\cot A = 2bc\cos A = b^2+c^2 - a^2$