C-09-Q214mediumsingle_mcqIf sin2A=1+cos2A\sin^2 A = 1 + \cos^2 Asin2A=1+cos2A, then A=?A = ?A=?a30∘30^\circ30∘b45∘45^\circ45∘c60∘60^\circ60∘d90∘90^\circ90∘ব্যাখ্যাThe condition gives sin2A−cos2A=1\sin^2A-\cos^2A=1sin2A−cos2A=1; combined with sin2A+cos2A=1\sin^2A+\cos^2A=1sin2A+cos2A=1 this forces cos2A=0\cos^2A=0cos2A=0, so A=90∘A=90^\circA=90∘.