Prove that:
sin2π6+cos2π3−tan2π4 =- 12
We have
L.H.S = sin2π6+cos2π3−tan2π4
= (12)2+(12)2−(1)2=14+14−1
= 1+1−44=−24=−12= R.H.S
sin2 (π8+A2)−sin2(π8−A2)=1√2 sin A