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Question-164530




Question Number 164530 by hoochhoch last updated on 18/Jan/22
Commented by Rasheed.Sindhi last updated on 18/Jan/22
sinα=−((12)/3)=−4⇒α∉R in Q#1
$$\mathrm{sin}\alpha=−\frac{\mathrm{12}}{\mathrm{3}}=−\mathrm{4}\Rightarrow\alpha\notin\mathbb{R}\:{in}\:{Q}#\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Jan/22
sinα=(3/5) ;((cotα+tanα  )/(cotα−tanα))=?  ((cotα+tanα  )/(cotα−tanα))=((((cosα )/(sinα ))+((sinα )/(cosα )))/(((cosα )/(sinα ))−((sinα )/(cosα ))))  =((cos^2 α+sin^2 α  )/(cos^2 α−sin^2 α))=(1/(1−2sin^2 α ))  =(1/(1−2((3/5))^2 ))=(1/((25−18)/(25)))=((25)/7)
$$\mathrm{sin}\alpha=\frac{\mathrm{3}}{\mathrm{5}}\:;\frac{\mathrm{cot}\alpha+\mathrm{tan}\alpha\:\:}{\mathrm{cot}\alpha−\mathrm{tan}\alpha}=? \\ $$$$\frac{\mathrm{cot}\alpha+\mathrm{tan}\alpha\:\:}{\mathrm{cot}\alpha−\mathrm{tan}\alpha}=\frac{\frac{\mathrm{cos}\alpha\:}{\mathrm{sin}\alpha\:}+\frac{\mathrm{sin}\alpha\:}{\mathrm{cos}\alpha\:}}{\frac{\mathrm{cos}\alpha\:}{\mathrm{sin}\alpha\:}−\frac{\mathrm{sin}\alpha\:}{\mathrm{cos}\alpha\:}} \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \alpha\:\:}{\mathrm{cos}^{\mathrm{2}} \alpha−\mathrm{sin}^{\mathrm{2}} \alpha}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \alpha\:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\frac{\mathrm{25}−\mathrm{18}}{\mathrm{25}}}=\frac{\mathrm{25}}{\mathrm{7}} \\ $$

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