## Task B-3.2.

Determine the acute angle $\alpha$ if $\operatorname{tg} \alpha=\frac{\left(1+\operatorname{tg} 1^{\circ}\right)\left(1+\operatorname{tg} 2^{\circ}\right)-2}{\left(1-\operatorname{tg} 1^{\circ}\right)\left(1-\operatorname{tg} 2^{\circ}\right)-2}$.
## First Solution.

Let's simplify the right side of the given equation. First, we will multiply the expressions in parentheses and simplify a bit.

$$
\begin{gathered}
\operatorname{tg} \alpha=\frac{\left(1+\operatorname{tg} 1^{\circ}\right)\left(1+\operatorname{tg} 2^{\circ}\right)-2}{\left(1-\operatorname{tg} 1^{\circ}\right)\left(1-\operatorname{tg} 2^{\circ}\right)-2}=\frac{1+\operatorname{tg} 1^{\circ}+\operatorname{tg} 2^{\circ}+\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-2}{1-\operatorname{tg} 1^{\circ}-\operatorname{tg} 2^{\circ}+\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-2} \\
=\frac{\operatorname{tg} 1^{\circ}+\operatorname{tg} 2^{\circ}+\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-1}{-\operatorname{tg} 1^{\circ}-\operatorname{tg} 2^{\circ}+\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-1}
\end{gathered}
$$

After simplification, divide the numerator and the denominator by $\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-1$. It follows that

$$
\operatorname{tg} \alpha=\frac{\frac{\operatorname{tg} 1^{\circ}+\operatorname{tg} 2^{\circ}}{\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-1}+1}{-\frac{\operatorname{tg} 1^{\circ}+\operatorname{tg} 2^{\circ}}{\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}-1}+1}
$$

Since $\operatorname{tg} 3^{\circ}=\frac{\operatorname{tg} 1^{\circ}+\operatorname{tg} 2^{\circ}}{1-\operatorname{tg} 1^{\circ} \cdot \operatorname{tg} 2^{\circ}}$, then

$$
\operatorname{tg} \alpha=\frac{1-\operatorname{tg} 3^{\circ}}{1+\operatorname{tg} 3^{\circ}}
$$

The right side of the given expression, using the fact that $\operatorname{tg} 45^{\circ}=1$, can be written as

$$
\operatorname{tg} \alpha=\frac{1-\operatorname{tg} 3^{\circ}}{1+\operatorname{tg} 3^{\circ}}=\frac{\operatorname{tg} 45^{\circ}-\operatorname{tg} 3^{\circ}}{1+\operatorname{tg} 45^{\circ} \cdot \operatorname{tg} 3^{\circ}}=\operatorname{tg}\left(45^{\circ}-3^{\circ}\right)=\operatorname{tg} 42^{\circ}
$$

Therefore, $\alpha=42^{\circ}$.