C-09-Q092mediumsingle_mcqIf secA+tanA=52\sec A + \tan A = \dfrac{5}{2}secA+tanA=25, then secA−tanA=?\sec A - \tan A = ?secA−tanA=?a52\dfrac{5}{2}25b25\dfrac{2}{5}52c111d−25-\dfrac{2}{5}−52ব্যাখ্যাSince sec2A−tan2A=1\sec^2 A - \tan^2 A = 1sec2A−tan2A=1, we have (secA+tanA)(secA−tanA)=1(\sec A+\tan A)(\sec A-\tan A) = 1(secA+tanA)(secA−tanA)=1. Therefore secA−tanA=15/2=25\sec A - \tan A = \dfrac{1}{5/2} = \dfrac{2}{5}secA−tanA=5/21=52.