C-09-Q172mediumsingle_mcqtan45∘⋅sin260∘−tan30∘⋅tan60∘=?\tan 45^\circ \cdot \sin^2 60^\circ - \tan 30^\circ \cdot \tan 60^\circ = ?tan45∘⋅sin260∘−tan30∘⋅tan60∘=?a14\dfrac{1}{4}41b−14-\dfrac{1}{4}−41c000d34\dfrac{3}{4}43ব্যাখ্যাHere tan45∘=1\tan 45^\circ=1tan45∘=1, sin260∘=34\sin^2 60^\circ=\tfrac34sin260∘=43, and tan30∘⋅tan60∘=13⋅3=1\tan 30^\circ\cdot\tan 60^\circ=\tfrac{1}{\sqrt3}\cdot\sqrt3=1tan30∘⋅tan60∘=31⋅3=1. So the value is 34−1=−14\tfrac34-1=-\tfrac1443−1=−41.