If A=15∘, find the value of 4cos2A⋅sin4A⋅tan3A−1.
Given: A=15∘;
Hence, 4cos2A⋅sin4A⋅tan3A−1 =4cos2(15∘)⋅sin4(15∘)⋅tan3(15∘)−1 =4cos30∘⋅sin60∘⋅tan45∘−1
=4√32×√32×1−1
=3−1
=2
If A = 15°
Find 4 cos 2A . sin 4A .tan 3A - 1