C-10-Q153mediumsingle_mcqtan30∘⋅tan60∘\tan 30^\circ \cdot \tan 60^\circtan30∘⋅tan60∘ equals—a111b3\sqrt{3}3c13\dfrac{1}{3}31d333ব্যাখ্যাSince tan30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}}tan30∘=31 and tan60∘=3\tan 60^\circ = \sqrt{3}tan60∘=3, their product is 13×3=1\frac{1}{\sqrt{3}} \times \sqrt{3} = 131×3=1. These two angles are complementary, and tanθ⋅tan(90∘−θ)=1\tan\theta \cdot \tan(90^\circ-\theta) = 1tanθ⋅tan(90∘−θ)=1.