Need Sine form of Cotangent equation

Question

$$(b^2 - c^2)\cot A + (c^2 - a^2)\cot B + (a^2 -b^2)\cot C=0$$

I want this equation to be in the Sine form.

Please help me with steps.

Thanks a lot

Answer 1

Assuming that $a,b,c$ are the sides of a Triangle

Using Law of sines $$T=(b^2-c^2)\cot A=4R^2(\sin^2B-\sin^2C)\frac{\cos A}{\sin A}$$

Again using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ or Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$

$\displaystyle\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$

But as $A+B+C=\pi$ and $\displaystyle\sin(\pi-x)=\sin x,\sin(B+C)=\sin(\pi-A)=\sin A$

$T$ becomes $\displaystyle 4R^2\sin A\sin(B-C)\frac{\cos A}{\sin A}$

As $\sin(B+C)=\sin A\ne0$ and $\cos(\pi-y)=-\cos y$

$T$ reduces to $\displaystyle 4R^2\sin(B-C)\{-\cos(B+C)\}=-2R^2\{2\sin(B-C)\cos(B+C)\}$

Using $\displaystyle2\sin B\cos A=\sin(A+B)-\sin(A-B), T=-2R^2(\underbrace{\sin2B-\sin2C}) $

Answer 2

Another way: Group the cotangent ratios

Considering one of them

$\displaystyle T_1=b^2(\cot A-\cot C)=\frac{b^2\sin(C-A)}{\sin A\sin C}$

As $\sin B=\sin\{\pi-(C+A)\}=\sin(C+A)$ and $\sin B\ne0$(why?)

$\displaystyle T_1=\frac{b^2\sin(C-A)\sin(C+A)}{\sin A\sin C\sin B}$

Using $\sin(x+y)\sin(x-y)$ formula(mentioned in the other answer)

$\displaystyle T_1=\frac{b^2(\sin^2C-\sin^2A)}{\sin A\sin C\sin B}\ \ \ \ (1)$

Now using sine Law, $\displaystyle \frac c{\sin C}=\frac b{\sin B}=2R$ where $R$ is the circum-radius

Set the values of $\sin C,\sin A$ in the numerator of $(1)$