Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 178181 by mathlove last updated on 13/Oct/22

((cosα cotα−sinα tanα)/(cscα secα))=?

$$\frac{{cos}\alpha\:{cot}\alpha−{sin}\alpha\:{tan}\alpha}{{csc}\alpha\:{sec}\alpha}=? \\ $$

Commented by Adedayo2000 last updated on 13/Oct/22

Recall cotα=(1/(tanα))=((cosα)/(sinα)), cscα=(1/(sinα)), secα=(1/(cosα))  =>(((cosα)(((cosα)/(sinα)))−(sinα)(((sinα)/(cosα))))/(((1/(sinα)))((1/(cosα)))))  =>((((cos^2 α)/(sinα))−((sin^2 α)/(cosα)))/(1/(sinαcosα))) =>(((cos^3 α−sin^3 α)/(sinαcosα))/(1/(sinαcosα)))  =>cos^3 α−sin^3 α   AδξδαΥo Δδξβαγo....

$${Recall}\:{cot}\alpha=\frac{\mathrm{1}}{{tan}\alpha}=\frac{{cos}\alpha}{{sin}\alpha},\:{csc}\alpha=\frac{\mathrm{1}}{{sin}\alpha},\:{sec}\alpha=\frac{\mathrm{1}}{{cos}\alpha} \\ $$$$=>\frac{\left({cos}\alpha\right)\left(\frac{{cos}\alpha}{{sin}\alpha}\right)−\left({sin}\alpha\right)\left(\frac{{sin}\alpha}{{cos}\alpha}\right)}{\left(\frac{\mathrm{1}}{{sin}\alpha}\right)\left(\frac{\mathrm{1}}{{cos}\alpha}\right)} \\ $$$$=>\frac{\frac{{cos}^{\mathrm{2}} \alpha}{{sin}\alpha}−\frac{{sin}^{\mathrm{2}} \alpha}{{cos}\alpha}}{\frac{\mathrm{1}}{{sin}\alpha{cos}\alpha}}\:=>\frac{\frac{{cos}^{\mathrm{3}} \alpha−{sin}^{\mathrm{3}} \alpha}{{sin}\alpha{cos}\alpha}}{\frac{\mathrm{1}}{{sin}\alpha{cos}\alpha}} \\ $$$$=>{cos}^{\mathrm{3}} \alpha−{sin}^{\mathrm{3}} \alpha \\ $$$$\:{A}\delta\xi\delta\alpha\Upsilon{o}\:\Delta\delta\xi\beta\alpha\gamma{o}.... \\ $$

Answered by Ar Brandon last updated on 13/Oct/22

((cosαcotα−sinαtanα)/(cscαsecα))=cos^3 α−sin^3 α

$$\frac{\mathrm{cos}\alpha\mathrm{cot}\alpha−\mathrm{sin}\alpha\mathrm{tan}\alpha}{\mathrm{csc}\alpha\mathrm{sec}\alpha}=\mathrm{cos}^{\mathrm{3}} \alpha−\mathrm{sin}^{\mathrm{3}} \alpha \\ $$

Answered by itzikhaim last updated on 13/Oct/22

=((cosα ((cosα)/(sinα))−sinα ((sinα)/(cosα)))/(cscα secα))  =((((cos^2 α)/(sinα))−((sin^2 α)/(cosα)))/(((1/(cosα)))((1/(sinα)))))=(((cos^2 α−sin^2 α)/(sinαcosα))/(1/(cosαsinα)))  =cos^2 α−sin^2 α  =cos(2α)

$$=\frac{{cos}\alpha\:\frac{{cos}\alpha}{{sin}\alpha}−{sin}\alpha\:\frac{{sin}\alpha}{{cos}\alpha}}{{csc}\alpha\:{sec}\alpha} \\ $$$$=\frac{\frac{{cos}^{\mathrm{2}} \alpha}{{sin}\alpha}−\frac{{sin}^{\mathrm{2}} \alpha}{{cos}\alpha}}{\left(\frac{\mathrm{1}}{{cos}\alpha}\right)\left(\frac{\mathrm{1}}{{sin}\alpha}\right)}=\frac{\frac{{cos}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \alpha}{\cancel{{sin}\alpha{cos}\alpha}}}{\frac{\mathrm{1}}{\cancel{{cos}\alpha{sin}\alpha}}} \\ $$$$={cos}^{\mathrm{2}} \alpha−{sin}^{\mathrm{2}} \alpha \\ $$$$={cos}\left(\mathrm{2}\alpha\right) \\ $$

Commented by Ar Brandon last updated on 13/Oct/22

((cos^2 α)/(sinα))−((sin^2 α)/(cosα))=((cos^3 α−sin^3 α)/(sinαcosα))

$$\frac{\mathrm{cos}^{\mathrm{2}} \alpha}{\mathrm{sin}\alpha}−\frac{\mathrm{sin}^{\mathrm{2}} \alpha}{\mathrm{cos}\alpha}=\frac{\mathrm{cos}^{\mathrm{3}} \alpha−\mathrm{sin}^{\mathrm{3}} \alpha}{\mathrm{sin}\alpha\mathrm{cos}\alpha} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com