## Task B-3.2.

Let $A=\frac{\cos 3+\cos 5}{\cos 3-\cos 5}$ and $B=\frac{\sin 5+\sin 7}{\cos 5-\cos 7}$.

Determine the real number $x$ such that $A+B x=0$.
## Solution.

$$
\begin{array}{cc}
A=\frac{\cos 3+\cos 5}{\cos 3-\cos 5}=\frac{2 \cos \frac{3+5}{2} \cos \frac{3-5}{2}}{-2 \sin \frac{3+5}{2} \sin \frac{3-5}{2}}=\frac{2 \cos 4 \cos (-1)}{-2 \sin 4 \sin (-1)}=\frac{2 \cos 4 \cos 1}{2 \sin 4 \sin 1}=\cot 4 \cot 1 & 2 \text { points } \\
B=\frac{\sin 5+\sin 7}{\cos 5-\cos 7}=\frac{2 \sin \frac{5+7}{2} \cos \frac{5-7}{2}}{-2 \sin \frac{5+7}{2} \sin \frac{5-7}{2}}=\frac{2 \sin 6 \cos (-1)}{-2 \sin 6 \sin (-1)}=\frac{\cos 1}{\sin 1}=\cot 1 & 2 \text { points }
\end{array}
$$

From $A+B x=0$ it follows that $\cot 4 \cot 1 + x \cot 4 = 0 \Rightarrow x = -\frac{\cot 4 \cdot \cot 1}{\cot 4} = -\cot 1$. 2 points