Gọi a,b là các số nguyên thỏa mãn

* Hướng dẫn giải

Chọn B.

Nhận xét: Nếu $A+B={45}^0$ thì $\left(1+\tan A\right)\left(1+\tan B\right)=2.$

Thật vậy:

$\left(1+\tan A\right)\left(1+\tan B\right)=\left(1+\tan A\right)\left[1+\tan \left({45}^0-A\right)\right]=\left(1+\tan A\right)\left[1+\frac{\tan {45}^0-\tan A}{1+\tan {45}^0.\tan A}\right]$

$=\left(1+\tan A\right)\left[1+\frac{1-\tan A}{1+\tan A}\right]=1+\tan A+1-\tan A=2.$

Khi đó:

$\left(1+\tan 1^0\right)\left(1+\tan 2^0\right)\left(1+\tan 3^0\right)\mathrm{...}\left(1+\tan {42}^0\right)\left(1+\tan {43}^0\right)=$

$$\begin{gathered} =\left(1+\tan 1^0\right)\left[\left(1+\tan 2^0\right)\left(1+\tan {43}^0\right)\right] \\ \left[\left(1+\tan 3^0\right)\left(1+\tan {42}^0\right)\right]\mathrm{...}\left[\left(1+\tan {22}^0\right)+\left(1+\tan {23}^0\right)\right] \end{gathered}$$

$=\left(1+\tan 1^0\right){.2}^{21}$. Suy ra $a=21,b=1.$

Vậy $P=a+b=22.$