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}23c111d000ব্যাখ্যাUsing A=60∘A=60^\circA=60∘, sinA=32\sin A=\dfrac{\sqrt3}{2}sinA=23 and cosA=12\cos A=\dfrac12cosA=21, so sin2A−cos2A=34−14=12\sin^2A-\cos^2A=\dfrac34-\dfrac14=\dfrac12sin2A−cos2A=43−41=21.