Question Number 212477 by Durganand last updated on 14/Oct/24
![](https://www.tinkutara.com/question/35184.png)
Answered by mehdee7396 last updated on 14/Oct/24
![((sin2acosa−sinacos2a)/(sin2asina)) =((sina)/(sin2asina))=(1/(sin2a)) ✓](https://www.tinkutara.com/question/Q212478.png)
$$\frac{{sin}\mathrm{2}{acosa}−{sinacos}\mathrm{2}{a}}{{sin}\mathrm{2}{asina}} \\ $$$$=\frac{{sina}}{{sin}\mathrm{2}{asina}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{a}}\:\:\checkmark \\ $$
Answered by A5T last updated on 14/Oct/24
![((cosA)/(sinA))−((cos2A)/(sin2A))=((2cos^2 A−(cos^2 A−sin^2 A))/(sin2A)) =((cos^2 A+sin^2 A)/(sin2A))=(1/(sin2A))](https://www.tinkutara.com/question/Q212484.png)
$$\frac{{cosA}}{{sinA}}−\frac{{cos}\mathrm{2}{A}}{{sin}\mathrm{2}{A}}=\frac{\mathrm{2}{cos}^{\mathrm{2}} {A}−\left({cos}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {A}\right)}{{sin}\mathrm{2}{A}} \\ $$$$=\frac{{cos}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} {A}}{{sin}\mathrm{2}{A}}=\frac{\mathrm{1}}{{sin}\mathrm{2}{A}} \\ $$