Question Number 216230 by MATHEMATICSAM last updated on 31/Jan/25
Answered by mahdipoor last updated on 31/Jan/25
$$\frac{{sec}^{\mathrm{2}} −{tan}}{{sec}^{\mathrm{2}} +{tan}}×\frac{{cos}^{\mathrm{2}} }{{cos}^{\mathrm{2}} }=\frac{\mathrm{1}−{sin}.{cos}}{\mathrm{1}+{sin}.{cos}}=\frac{\mathrm{2}−{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}+{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\frac{{A}}{{B}}\:\:\:\:\:\mathrm{1}<{A},{B}<\mathrm{3}\:\Rightarrow\:\:{min}\:,\:{max}=\frac{\mathrm{1}}{\mathrm{3}}\:,\:\frac{\mathrm{3}}{\mathrm{1}} \\ $$