C-09-Q182mediumsingle_mcqIf A=45∘A = 45^\circA=45∘, 1−tan2A1+tan2A=?\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = ?1+tan2A1−tan2A=?a000b111c−1-1−1d12\dfrac{1}{2}21ব্যাখ্যাThis is the identity 1−tan2A1+tan2A=cos2A\frac{1-\tan^2 A}{1+\tan^2 A}=\cos 2A1+tan2A1−tan2A=cos2A. For A=45∘A=45^\circA=45∘, cos90∘=0\cos 90^\circ=0cos90∘=0.