The correct option is C 1√3
Given: cot 12∘× cot 38∘× cot 52∘× cot 60∘× cot 78∘
On rearranging the given equation we get,
=(cot 12∘×cot 78∘)×(cot 38∘×cot 52∘)×cot 60∘
={cot 12∘×cot(90∘−12∘)}{cot 38∘×cot(90∘−38∘)}×cot 60∘
We know that, cot (90∘−θ)=tan θ.
Hence, cot (90∘−12∘)=tan 12∘,
cot (90∘−38∘)=tan 38∘ and cot 60∘=1√3
On substituting these values we get,
=(cot 12∘×tan 12∘)×(cot 38∘×tan 38∘)×1√3
=1×1×1√3=1√3.....[∵cot θ×tan θ=1] .
The value of the expression
cot 12∘× cot 38∘× cot 52∘× cot 60∘× cot 78∘=1√3.