Question

IMPOSSIBLE MATH QUESTION

Answer 1

Well u can also do this without a calculator
take $tan 1=tan(45-44)=\frac{1-tan44}{1+tan 44}$ 
then 
$(1+tan1)=(1+ \frac{1-tan44}{1+tan 44})=\frac{2}{1+tan 44}$
therefore similarly
$(1+tan 1)(1+tan 2)...(1+tan 22)= \frac{2}{1+tan 44} \frac{2}{1+tan 43}\frac{2}{1+tan 42}..\frac{2}{1+tan 23}$
$= \frac{2^{22}}{(1+tan 44)(1+tan 43)(1+tan 42)...(1+tan 23)}$
then $(1+tan 1)...(1+tan 45)= \frac{2^{22}}{(1+tan 44)(1+tan 43)(1+tan 42)...(1+tan 23)}(1+tan 23)(1+tan 24)...(1+tan 45)$
$=2^{22}(1+tan45)=2^{23}=8388608$

Answer 2

(1+tan 1)(1+ tan 2)...(1+tan 45)=(1+0.017455)(1+0.03492)...(1+1)=8388608.11