Problem:
Evaluate
log10(tan1∘)+log10(tan2∘)+log10(tan3∘)+⋯+log10(tan88∘)+log10(tan89∘)
Answer Choices:
A. 0
B. 21log10(213)
C. 21log102
D. 1
E. none of these
Solution:
Using loga+logb=logab repeatedly, we find that the sum is
P=log10[(tan1∘)(tan2∘)⋯(tan45∘)⋯(tan89∘)]
Moreover, (tan1∘)(tan89∘)=1,(tan2∘)(tan88∘)=1, and so on, because tanθtan(90−θ)=tanθcotθ=1 for all θ at which both tanθ and cotθ are defined. Thus
P=log10(tan45∘)=log101=0