Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 167929 by Huy last updated on 29/Mar/22

prove:          (((a^2 +b^2 )sinαcosα−ab)/((a^2 +b^2 )cos^2 α−b^2 ))=((asinα−bcosα)/(acosα+bsinα))

$${prove}:\: \\ $$$$\:\:\:\:\:\:\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\mathrm{sin}\alpha\mathrm{cos}\alpha−{ab}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\mathrm{cos}^{\mathrm{2}} \alpha−{b}^{\mathrm{2}} }=\frac{{a}\mathrm{sin}\alpha−{b}\mathrm{cos}\alpha}{{a}\mathrm{cos}\alpha+{b}\mathrm{sin}\alpha} \\ $$

Answered by som(math1967) last updated on 29/Mar/22

((a^2 sinαcosα+b^2 sinαcosα−absin^2 α−abcos^2 α)/(a^2 cos^2 α−b^2 (1−cos^2 α)))  =((asinα(acosα−bsinα)−bcosα(acosα−bsinα))/(a^2 cos^2 α−b^2 sin^2 α))  =(((acosα−bsinα)(asinα−bcosα))/((acosα+bsinα)(acosα−bsinα)))  =((asinα−bcosα)/(acosα+bsinα))    (proved)

$$\frac{{a}^{\mathrm{2}} {sin}\alpha{cos}\alpha+{b}^{\mathrm{2}} {sin}\alpha{cos}\alpha−{absin}^{\mathrm{2}} \alpha−{abcos}^{\mathrm{2}} \alpha}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \alpha−{b}^{\mathrm{2}} \left(\mathrm{1}−{cos}^{\mathrm{2}} \alpha\right)} \\ $$$$=\frac{{asin}\alpha\left({acos}\alpha−{bsin}\alpha\right)−{bcos}\alpha\left({acos}\alpha−{bsin}\alpha\right)}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \alpha−{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \alpha} \\ $$$$=\frac{\left({acos}\alpha−{bsin}\alpha\right)\left({asin}\alpha−{bcos}\alpha\right)}{\left({acos}\alpha+{bsin}\alpha\right)\left({acos}\alpha−{bsin}\alpha\right)} \\ $$$$=\frac{{asin}\alpha−{bcos}\alpha}{{acos}\alpha+{bsin}\alpha} \\ $$$$\:\:\left({proved}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com