The correct option is C 229
Given : (√3+tan1∘)(√3+tan2∘)(√3+tan3∘)⋯⋯(√3+tan28∘)(√3+tan29∘)
We know that
√3+tan(30∘−x)=√3+tan30∘−tanx1+tan30∘⋅tanx=√3+1−√3tanx√3+tanx=4√3+tanx
Now,
(√3+tan1∘)(√3+tan29∘)=4√3+tan29∘×(√3+tan29∘)=4
Similarly,
(√3+tan2∘)(√3+tan28∘)=4
So,
(√3+tan1∘)(√3+tan2∘)(√3+tan3∘)⋯(√3+tan28∘)(√3+tan29∘)=414⋅(√3+tan15∘)=414⋅(√3+√3−1√3+1)=414⋅(2+2√3√3+1)=229