Question Number 178181 by mathlove last updated on 13/Oct/22
$$\frac{{cos}\alpha\:{cot}\alpha−{sin}\alpha\:{tan}\alpha}{{csc}\alpha\:{sec}\alpha}=? \\ $$
Commented by Adedayo2000 last updated on 13/Oct/22
$${Recall}\:{cot}\alpha=\frac{\mathrm{1}}{{tan}\alpha}=\frac{{cos}\alpha}{{sin}\alpha},\:{csc}\alpha=\frac{\mathrm{1}}{{sin}\alpha},\:{sec}\alpha=\frac{\mathrm{1}}{{cos}\alpha} \\ $$$$=>\frac{\left({cos}\alpha\right)\left(\frac{{cos}\alpha}{{sin}\alpha}\right)−\left({sin}\alpha\right)\left(\frac{{sin}\alpha}{{cos}\alpha}\right)}{\left(\frac{\mathrm{1}}{{sin}\alpha}\right)\left(\frac{\mathrm{1}}{{cos}\alpha}\right)} \\ $$$$=>\frac{\frac{{cos}^{\mathrm{2}} \alpha}{{sin}\alpha}−\frac{{sin}^{\mathrm{2}} \alpha}{{cos}\alpha}}{\frac{\mathrm{1}}{{sin}\alpha{cos}\alpha}}\:=>\frac{\frac{{cos}^{\mathrm{3}} \alpha−{sin}^{\mathrm{3}} \alpha}{{sin}\alpha{cos}\alpha}}{\frac{\mathrm{1}}{{sin}\alpha{cos}\alpha}} \\ $$$$=>{cos}^{\mathrm{3}} \alpha−{sin}^{\mathrm{3}} \alpha \\ $$$$\:{A}\delta\xi\delta\alpha\Upsilon{o}\:\Delta\delta\xi\beta\alpha\gamma{o}…. \\ $$
Answered by Ar Brandon last updated on 13/Oct/22
$$\frac{\mathrm{cos}\alpha\mathrm{cot}\alpha−\mathrm{sin}\alpha\mathrm{tan}\alpha}{\mathrm{csc}\alpha\mathrm{sec}\alpha}=\mathrm{cos}^{\mathrm{3}} \alpha−\mathrm{sin}^{\mathrm{3}} \alpha \\ $$
Answered by itzikhaim last updated on 13/Oct/22
$$=\frac{{cos}\alpha\:\frac{{cos}\alpha}{{sin}\alpha}−{sin}\alpha\:\frac{{sin}\alpha}{{cos}\alpha}}{{csc}\alpha\:{sec}\alpha} \\ $$$$=\frac{\frac{{cos}^{\mathrm{2}} \alpha}{{sin}\alpha}−\frac{{sin}^{\mathrm{2}} \alpha}{{cos}\alpha}}{\left(\frac{\mathrm{1}}{{cos}\alpha}\right)\left(\frac{\mathrm{1}}{{sin}\alpha}\right)}=\frac{\frac{{cos}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \alpha}{\cancel{{sin}\alpha{cos}\alpha}}}{\frac{\mathrm{1}}{\cancel{{cos}\alpha{sin}\alpha}}} \\ $$$$={cos}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \alpha \\ $$$$={cos}\left(\mathrm{2}\alpha\right) \\ $$
Commented by Ar Brandon last updated on 13/Oct/22
$$\frac{\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{sin}\alpha}−\frac{\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{cos}\alpha}=\frac{\mathrm{cos}^{\mathrm{3}} \alpha−\mathrm{sin}^{\mathrm{3}} \alpha}{\mathrm{sin}\alpha\mathrm{cos}\alpha} \\ $$