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 \((1+tan 1)...(1+tan 45)= \frac{2^{22}}{(1+tan 44)(1+tan 43)(1+tan 42)...(1+tan 23)}(1+tan 23)(1+tan 24)...(1+tan 45)\)
\(=2^{22}(1+tan45)=2^{23}=8388608\)
