Prove $b^{2} (\cot A + \cot B) = c^{2}(\cot A + \cot C)$ In any triangle  $ABC $ prove that  $$b^{2} (\cot A + \cot B) = c^{2}(\cot A + \cot C)$$
How we can prove this trigonometric identity.  I tried many ways and use the other well known identity but it wasn't work. My question is how we can prove this trigonometric identity?. Any hint will help  Thanks.

There is a nice proof of this trig identity. 
Notice, in right $\triangle ABC$  $(A+B+C=180^\circ)$ we know from sine rule $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$$ 
$$\implies \ a=k\sin A, \  b=k\sin B, \ c=k\sin C, $$
Now, we have $$LHS=b^2(\cot A+\cot B)$$
$$=(k\sin B)^2\left(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}\right)$$
$$=k^2\sin^2B \left(\frac{\sin A\cos B+\cos A\sin B}{\sin A\sin B}\right)$$
$$=k^2 \frac{\sin B\sin (A+B)}{\sin A}$$
 $$=k^2 \frac{\sin (180^\circ-(A+C))\sin (180^\circ-C)}{\sin A}$$
 $$=k^2 \frac{\sin (A+C)\sin C}{\sin A}$$
$$=k^2\sin^2 C \left(\frac{\sin A\cos C+\cos A\sin C}{\sin A\sin C}\right)$$
 $$=(k\sin C)^2 \left(\frac{\cos C}{\sin C}+\frac{\cos A}{\sin A}\right)$$
setting the value $k\sin C=c$ from (1) $$=\color{}{(c)^2(\cot C+\cot A)}$$
$$LHS=\color{red}{c^2(\cot A+\cot C)}=RHS$$
