<s>[INST] Prove the given inequality $$\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $$
$$\implies \frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
Given if $A>B>C$ and $A+B+C=180^\circ$
Is that implication correct if not then please correct it otherwise try to solve the first inequality.
This is not the original problem, but this problem arose when I was solving another trigonometric equation.
Now if anybody prove or disprove the above inequality then that will also be the solution of my problem.
 [/INST] The original inequality is $\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $
And we have to prove this $\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$.
Now we have to interchange only the positions of $tan(B-C)$ and $sin^{2}B$. $sin^{2}B$ is always positive since it is a square. We need to care only about $tan(B-C)$.
If $tan(B-C)<0$ then $(B-C)>90^\circ \implies B>90^\circ+C$. 
$C$ is a non-zero angle. Therefore $B>90^\circ$ and $A>B>90^\circ$. 
Consider the fact that $A+B+C=180^\circ$. This means that the angles A, B , C form the angles of a triangle. But we know that there cannot exist two obtuse angles in a triangle. Therefore our supposition is wrong and hence $B-C<90^\circ$ which leads to $tan(B-C)>0$.
Now we can interchange $tan(B-C)$ and $sin^{2}B$ without changing the inequality sign. Hence we have the given inequality $$\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
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