C-09-Q209mediumsingle_mcqIn the same triangle, sin2A−cos2A=?\sin^2 A - \cos^2 A = ?sin2A−cos2A=?a12\dfrac{1}{2}21b32\dfrac{\sqrt{3}}{2}ব্যাখ্যাUsing A=60∘A=60^\circA=60∘, sinA=32\sin A=\dfrac{\sqrt3}{2}sinA=2 and , so .