Question

(Sin 30° +Cos 30')-( Sin 60° +Cos 60°) =​

Answer 1

Answer:

= 0

Step-by-step explanation:

$sin30^0=cos60^0=\frac{1}{2} \\\\ cos30^0=sin60^0=\frac{\sqrt{3}}{2} \\\\ => (Sin 30^0 +Cos 30^0)-( Sin 60^0 +Cos 60^0) \\\\=>\frac{1}{2} +\frac{\sqrt{3} }{2} -[\frac{\sqrt{3} }{2} +\frac{1}{2} ] \\\\ =>\frac{1}{2} +\frac{\sqrt{3} }{2} -\frac{\sqrt{3} }{2} -\frac{1}{2} \\\\ =>0$