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Question-14991

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Question Number 14991 by tawa tawa last updated on 06/Jun/17
Answered by ajfour last updated on 06/Jun/17
=((tan θ)/((1−(1/(tan θ)))))+(1/(tan θ(1−tan θ))) =((tan^2 θ)/(tan θ−1))−(1/(tan θ(tan θ−1))) =((tan^3 θ−1)/(tan θ(tan θ−1))) =((tan^2 θ+1+tan θ)/(tan θ)) (if tan θ≠1) =((sec^2 θ+tan θ)/(tan θ)) =((sec^2 θ)/(tan θ))+1 =(1/((cos^2 θ)(((sin θ)/(cos θ)))))+1 =(1/(sin θcos θ))+1 =cosec θsec θ+1 .
$$\:\:\:\:\:=\frac{\mathrm{tan}\:\theta}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right)}+\frac{\mathrm{1}}{\mathrm{tan}\:\theta\left(\mathrm{1}−\mathrm{tan}\:\theta\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:\theta−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta\left(\mathrm{tan}\:\theta−\mathrm{1}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{tan}\:^{\mathrm{3}} \theta−\mathrm{1}}{\mathrm{tan}\:\theta\left(\mathrm{tan}\:\theta−\mathrm{1}\right)}\: \\ $$$$\:\:\:\:=\frac{\mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{1}+\mathrm{tan}\:\theta}{\mathrm{tan}\:\theta}\:\:\:\:\:\:\left({if}\:\mathrm{tan}\:\theta\neq\mathrm{1}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{sec}\:^{\mathrm{2}} \theta+\mathrm{tan}\:\theta}{\mathrm{tan}\:\theta}\:=\frac{\mathrm{sec}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:\theta}+\mathrm{1} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{cos}\:^{\mathrm{2}} \theta\right)\left(\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)}+\mathrm{1} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}+\mathrm{1} \\ $$$$\:\:\:\:=\mathrm{cosec}\:\theta\mathrm{sec}\:\theta+\mathrm{1}\:. \\ $$
Commented by tawa tawa last updated on 06/Jun/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by RasheedSoomro last updated on 06/Jun/17
((tanθ)/(1−cotθ))+((cotθ)/(1−tanθ))=1+secθcscθ LHS:(((sinθ)/(cosθ))/(1−((cosθ)/(sinθ))))+(((cosθ)/(sinθ))/(1−((sinθ)/(cosθ)))) =(((sinθ)/(cosθ))/((sinθ−cosθ)/(sinθ)))+(((cosθ)/(sinθ))/((cosθ−sinθ)/(cosθ))) = ((sinθ)/(cosθ))×((sinθ)/(sinθ−cosθ))+((cosθ)/(sinθ))×((cosθ)/(cosθ−sinθ)) =((sin^2 θ)/(cosθ(sinθ−cosθ)))+((cos^2 θ)/(sinθ(cosθ−sinθ))) =(1/(sinθ−cosθ))(((sin^2 θ)/(cosθ))−((cos^2 θ)/(sinθ))) =(1/(sinθ−cosθ))(((sin^3 θ−cos^3 θ)/(sinθcosθ))) =(1/(sinθ−cosθ))×(((sinθ−cosθ)(sin^2 θ+sinθcosθ+cos^2 θ))/(sinθcosθ)) =((sin^2 θ+sinθcosθ+cos^2 θ)/(sinθcosθ)) =((1+sinθcosθ)/(sinθcosθ)) =(1/(sinθcosθ))+((sinθcosθ)/(sinθcosθ)) =1+(1/(cosθ))×(1/(sinθ)) =1+secθ cscθ=RHS
$$\frac{\mathrm{tan}\theta}{\mathrm{1}−\mathrm{cot}\theta}+\frac{\mathrm{cot}\theta}{\mathrm{1}−\mathrm{tan}\theta}=\mathrm{1}+\mathrm{sec}\theta\mathrm{csc}\theta \\ $$$$\mathrm{LHS}:\frac{\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}}{\mathrm{1}−\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}}+\frac{\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}}{\mathrm{1}−\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}} \\ $$$$\:\:\:\:=\frac{\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}}{\frac{\mathrm{sin}\theta−\mathrm{cos}\theta}{\mathrm{sin}\theta}}+\frac{\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}}{\frac{\mathrm{cos}\theta−\mathrm{sin}\theta}{\mathrm{cos}\theta}} \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}×\frac{\mathrm{sin}\theta}{\mathrm{sin}\theta−\mathrm{cos}\theta}+\frac{\mathrm{cos}\theta}{\mathrm{sin}\theta}×\frac{\mathrm{cos}\theta}{\mathrm{cos}\theta−\mathrm{sin}\theta} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{cos}\theta\left(\mathrm{sin}\theta−\mathrm{cos}\theta\right)}+\frac{\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{sin}\theta\left(\mathrm{cos}\theta−\mathrm{sin}\theta\right)} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\theta−\mathrm{cos}\theta}\left(\frac{\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{cos}\theta}−\frac{\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{sin}\theta}\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\theta−\mathrm{cos}\theta}\left(\frac{\mathrm{sin}^{\mathrm{3}} \theta−\mathrm{cos}^{\mathrm{3}} \theta}{\mathrm{sin}\theta\mathrm{cos}\theta}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\theta−\mathrm{cos}\theta}×\frac{\left(\mathrm{sin}\theta−\mathrm{cos}\theta\right)\left(\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{sin}\theta\mathrm{cos}\theta+\mathrm{cos}^{\mathrm{2}} \theta\right)}{\mathrm{sin}\theta\mathrm{cos}\theta} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{sin}\theta\mathrm{cos}\theta+\mathrm{cos}^{\mathrm{2}} \theta}{\mathrm{sin}\theta\mathrm{cos}\theta} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{sin}\theta\mathrm{cos}\theta}{\mathrm{sin}\theta\mathrm{cos}\theta} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}\theta\mathrm{cos}\theta}+\frac{\mathrm{sin}\theta\mathrm{cos}\theta}{\mathrm{sin}\theta\mathrm{cos}\theta} \\ $$$$\:\:\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{cos}\theta}×\frac{\mathrm{1}}{\mathrm{sin}\theta} \\ $$$$\:\:\:\:\:=\mathrm{1}+\mathrm{sec}\theta\:\mathrm{csc}\theta=\mathrm{RHS} \\ $$$$ \\ $$
Commented by tawa tawa last updated on 06/Jun/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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