Suppose that $0^{\circ}<A<90^{\circ}$ and $0^{\circ}<B<90^{\circ}$ and

$$
\left(4+\tan ^{2} A\right)\left(5+\tan ^{2} B\right)=\sqrt{320} \tan A \tan B
$$

Determine all possible values of $\cos A \sin B$.

## PART B

For each question in Part B, your solution must be well-organized and contain words of explanation or justification. Marks are awarded for completeness, clarity, and style of presentation. A correct solution, poorly presented, will not earn full marks.
Solution 1

When $0^{\circ}0$ and $\tan B>0$.

For all real numbers $x$ and $y$, we know that $(x-y)^{2} \geq 0$ with equality exactly when $x=y$.

Rearranging, we obtain $x^{2}-2 x y+y^{2} \geq 0$ and $x^{2}+y^{2} \geq 2 x y$ with equality exactly when $x=y$.

Setting $x=2$ and $y=\tan A$, we obtain $4+\tan ^{2} A \geq 2(2)(\tan A)$.

Thus, $4+\tan ^{2} A \geq 4 \tan A$ with equality exactly when $\tan A=2$.

Setting $x=\sqrt{5}$ and $y=\tan B$, we obtain $5+\tan ^{2} B \geq 2(\sqrt{5})(\tan B)$.

Thus, $5+\tan ^{2} B \geq 2 \sqrt{5} \tan B$ with equality exactly when $\tan B=\sqrt{5}$.

Since $4+\tan ^{2} A \geq 4 \tan A$ and $5+\tan ^{2} B \geq 2 \sqrt{5} \tan B$, then

$$
\left(4+\tan ^{2} A\right)\left(5+\tan ^{2} B\right) \geq(4 \tan A)(2 \sqrt{5} \tan B)
$$

since all quantities are positive.

For equality to occur in this inequality, we need equality to occur in both of the component inequalities. (If one inequality is actually " $>$ ", then the product becomes " $>$ ".) In other words, $\left(4+\tan ^{2} A\right)\left(5+\tan ^{2} B\right)=(4 \tan A)(2 \sqrt{5} \tan B)$ exactly when $\tan A=2$ and $\tan B=\sqrt{5}$.

We note that

$$
(4 \tan A)(2 \sqrt{5} \tan B)=8 \sqrt{5} \tan A \tan B=\sqrt{64 \cdot 5} \tan A \tan B=\sqrt{320} \tan A \tan B
$$

Therefore, $\left(4+\tan ^{2} A\right)\left(5+\tan ^{2} B\right)=\sqrt{320} \tan A \tan B$ exactly when $\tan A=2$ and $\tan B=\sqrt{5}$.

We now need to determine the value of $\cos A \sin B$.

Since $0^{\circ}0$, then $\cos A=\frac{1}{\sqrt{5}}$.

When $\tan B=\sqrt{5}$, we obtain $\frac{\sin B}{\cos B}=\sqrt{5}$ and so $\frac{\sin B}{\sqrt{1-\sin ^{2} B}}=\sqrt{5}$.

This gives $\sin ^{2} B=5-5 \sin ^{2} B$ or $6 \sin ^{2} B=5$.

Since $\sin B>0$, then $\sin B=\frac{\sqrt{5}}{\sqrt{6}}$.

Finally, this gives $\cos A \sin B=\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{6}}=\frac{1}{\sqrt{6}}$.

Solution 2

Let $X=\tan A$ and $Y=\tan B$.

The given equation becomes successively:

$$
\begin{aligned}
\left(4+X^{2}\right)\left(5+Y^{2}\right) & =\sqrt{320} X Y \\
X^{2} Y^{2}+5 X^{2}+4 Y^{2}+20 & =8 \sqrt{5} X Y \\
X^{2} Y^{2}-8 \sqrt{5} X Y+5 X^{2}+4 Y^{2}+20 & =0 \\
X^{2} Y^{2}-2(2 \sqrt{5}) X Y+20+5 X^{2}+4 Y^{2}-4 \sqrt{5} X Y & =0 \\
X^{2} Y^{2}-2(2 \sqrt{5}) X Y+(2 \sqrt{5})^{2}+5 X^{2}+4 Y^{2}-4 \sqrt{5} X Y & =0 \\
(X Y-2 \sqrt{5})^{2}+5 X^{2}-4 \sqrt{5} X Y+4 Y^{2} & =0 \\
(X Y-2 \sqrt{5})^{2}+(\sqrt{5} X)^{2}-2(\sqrt{5} X)(2 Y)+(2 Y)^{2} & =0 \\
(X Y-2 \sqrt{5})^{2}+(\sqrt{5} X-2 Y)^{2} & =0
\end{aligned}
$$

For the sum of squares to be 0 , each part must equal 0 .

Therefore, $X Y=2 \sqrt{5}$ and $\sqrt{5} X=2 Y$.

From the first equation, $X(2 Y)=4 \sqrt{5}$ and so $X(\sqrt{5} X)=4 \sqrt{5}$ which gives $X^{2}=4$ or $X= \pm 2$.

Since $0^{\circ}0$ and so $X=\tan A=2$.

Since $2 Y=\sqrt{5} X$, then $Y=\tan B=\sqrt{5}$.

We can then proceed as in Solution 1 to obtain $\cos A \sin B=\frac{1}{\sqrt{6}}$.

## Part B