Question

Sin 70°cos 10°-cos 70°sin 10=√3/2

Answer 1

Let a =70° and b= 10° (and a-b=70-10 =60)
We have the following trigonometric identity:

sin(a-b) = sin(a).cos(b)-sin(b).cos(a) OR:
sin(70-10) = sin70.cos10 - sin10.cos70
But sin(70-10) = sin(60) and we know that sin(60°) =(√3)/2