Question Number 212477 by Durganand last updated on 14/Oct/24
Answered by mehdee7396 last updated on 14/Oct/24
$$\frac{{sin}\mathrm{2}{acosa}−{sinacos}\mathrm{2}{a}}{{sin}\mathrm{2}{asina}} \\ $$$$=\frac{{sina}}{{sin}\mathrm{2}{asina}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{a}}\:\:\checkmark \\ $$
Answered by A5T last updated on 14/Oct/24
$$\frac{{cosA}}{{sinA}}−\frac{{cos}\mathrm{2}{A}}{{sin}\mathrm{2}{A}}=\frac{\mathrm{2}{cos}^{\mathrm{2}} {A}−\left({cos}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {A}\right)}{{sin}\mathrm{2}{A}} \\ $$$$=\frac{{cos}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {A}}{{sin}\mathrm{2}{A}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{A}} \\ $$