Creative Solution for IMO Trigonometry Problem

Let $A,\,B$ be acute angles such that $\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$.

Find the greatest possible value of $\tan B$.

This blog post is to highlight the fact that if we're creative enough, we can avoid the tedious calculus method to look for the maximal of $\tan B$.

You have to be aware of a few things as well:

1.

When $B$ is an acute angle and if $\sin B\le \dfrac{m}{n}$, then  $\tan B\le \dfrac{m}{\sqrt{n^2-m^2}}$ must be true.

In other words, we obtain the maximal of $\tan B$ if we have obtained the maximal of $\sin B$.

IMO (Hong Kong) Trigonometric Problem (Modified)

Let $A,\,B$ be acute angles such that $\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$.

Find the greatest possible value of $\tan B$.

This is a fun IMO problem, since it has many ways (all are nothing less than remarkable) to approach it and without any further ado, I will post with the first approach here:

$\tan B=2015\sin A \cos A-2015\sin^2 A \tan B$

$\tan B(1+2015\sin^2 A )=2015\sin A \cos A$