<s>[INST] 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.
 [/INST] 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) $$=\color{}{(c)^2(\cot C+\cot A)}$$
$$LHS=\color{red}{c^2(\cot A+\cot C)}=RHS$$
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