If 1 + tan 1 ° 1 + tan 2 ° … 1 + tan 45 ° = 2 n , then n =

Consider $A+B=45°\Rightarrow \tan \left(A+B\right)=1\Rightarrow \tan A+\tan B=1-\tan A\tan B$

$\Rightarrow 1+\tan A+\tan B+\tan A\tan B=2$

$\Rightarrow (1+\tan A)(1+\tan B)=2$

So, $\left(1+\tan 1°\right)\left(1+\tan 44°\right)=\left(1+\tan 2°\right)\left(1+\tan 43°\right)=....=\left(1+\tan 22°\right)\left(1+\tan 23°\right)=2$

Thus, $\left(1+\tan 1°\right)\left(1+\tan 2°\right)......\left(1+\tan 43°\right)\left(1+\tan 44°\right)=2^{22}$

Now, $1+\tan 45°=1+1=2$

So, $(1+\tan 1^o)\left(1+\tan 2^o\right)\dots .\left(1+\tan 44°\right)\left(1+\tan 45°\right)=2^{23}$

$\Rightarrow n=23$