Correct option is B
Use identity:tan(90∘−θ)=cotθSo:∙ tan80∘=cot10∘∙ tan70∘=cot20∘Therefore:tan10∘⋅tan20∘⋅tan60∘⋅tan70∘⋅tan80∘=tan10∘⋅tan20∘⋅tan60∘⋅cot20∘⋅cot10∘Now notice:tan10∘⋅cot10∘=1,tan20∘⋅cot20∘=1So the expression becomes:1⋅1⋅tan60∘=tan60∘=3