## Task A-3.3.

Calculate the product $\left(1+\operatorname{tg} 1^{\circ}\right)\left(1+\operatorname{tg} 2^{\circ}\right) \ldots\left(1+\operatorname{tg} 45^{\circ}\right)$.
## Solution.

From the addition formula for tangent, we have $\operatorname{tg}(\alpha+\beta)(1-\operatorname{tg} \alpha \operatorname{tg} \beta)=\operatorname{tg} \alpha+\operatorname{tg} \beta$.

If $\alpha+\beta=45^{\circ}$, then $1=\operatorname{tg} \alpha \operatorname{tg} \beta+\operatorname{tg} \alpha+\operatorname{tg} \beta$.

Thus,

$$
(1+\operatorname{tg} \alpha)\left(1+\operatorname{tg}\left(45^{\circ}-\alpha\right)\right)=1+\operatorname{tg} \alpha+\operatorname{tg}\left(45^{\circ}-\alpha\right)+\operatorname{tg} \alpha \operatorname{tg}\left(45^{\circ}-\alpha\right)=2
$$

for $\alpha=1^{\circ}, 2^{\circ}, \ldots, 44^{\circ}$.

5 points

In the desired product, the first 44 parentheses can be grouped into 22 pairs, with the product of two parentheses in each pair being equal to 2.

The value of the last parenthesis is also 2, so the desired product is equal to $2^{23}$.