20. Let $x, y$ be acute angles such that $\sin y=2005 \cos (x+y) \sin x$. Find the greatest possible value of $\tan y$.
(3 marks)
Let $x, y$ be acute angles such that $\sin y=2005 \cos (x+y) \sin x$. Find the greatest possible value of $\tan y$.
20. $\frac{2005 \sqrt{2006}}{4012}$
20. Applying the product-to-sum formula, we have
$$
\begin{aligned}
\sin B & =2005 \cos (A+B) \sin A \\
& =\frac{2005}{2}[\sin (2 A+B) - \sin (-B)] \\
\frac{2007}{2} \sin B & =\frac{2005}{2} \sin (2 A+B) \\
\sin B & =\frac{2005}{2007} \sin (2 A+B)
\end{aligned}
$$

Hence $\sin B \leq \frac{2005}{2007}$, and so $\tan B \leq \frac{2005}{\sqrt{2007^{2}-2005^{2}}}=\frac{2005}{2 \sqrt{2006}}=\frac{2005 \sqrt{2006}}{4012}$. Equality is possible when $\sin (2 A+B)=1$, i.e. when $2 A+B$ is a right angle.