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 54536 by 951172235v last updated on 05/Feb/19

If A,B,C are angles of a triangle show that tan^(−1) (cot Acot B)+tan^(−1) (cot Bcot C)+tan^(−1) (cot Ccot A) = tan^(−1) {1+((8cos Acos Bcos C)/(sin^2 2A+sin^2 2B+sin^2 2C))}

$$\mathrm{If}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{are}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\mathrm{Acot}\:\mathrm{B}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\mathrm{Bcot}\:\mathrm{C}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\mathrm{Ccot}\:\mathrm{A}\right) \\ $$$$=\:\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{1}+\frac{\mathrm{8cos}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2B}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2C}}\right\} \\ $$

Answered by 951172235v last updated on 08/Feb/19

A+B+C =Λ^− tan A+tan B+tan C = tan Atan Btan tan C tan α =cot Acot B tan β =cot Bcot C tan γ =cot Ccot A tan α+tan β+tan γ =1 tan (α+β+γ) =((Σtan α−tan αtan βtan γ)/(1−Σtan αtan β)) = ((1−(cot Acot Bcot C)^2 )/(1−cot Acot Bcot C(cot A+cot B+cot C))) =((tan Atan Btan C−cot Acot Bcot C)/(Σtan A−Σcot A)) = ((tan Atan Btan C−cot Acot Bcot C)/(Σ(((−cos 2A)/(sin 2A))))) = ((8(sin Asin Bsin C)^2 −8(cos Acos Bcos C)^2 )/(−Σsin 2A(sin 2Bcos 2C+sin 2Ccos 2B))) = (((1/2){(Σsin 2A)^2 −[1+Σcos 2A]^2 })/(Σsin^2 2A)) =(((1/2){2Σsin^2 A −4Σcos 2A−4})/(Σsin^2 2A)) = ((Σsin^2 2A+8cos Acos Bcos C)/(Σsin^2 2A)) =1+ ((8cos Acos Bcos C)/(Σsin^2 2A)) α+β+γ =tan^(−1) {1+((8cos Acos Bcos C)/(Σsin^2 2A))} ans.

$$\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\overset{−} {\Lambda}\:\: \\ $$$$\mathrm{tan}\:\mathrm{A}+\mathrm{tan}\:\mathrm{B}+\mathrm{tan}\:\mathrm{C}\:=\:\mathrm{tan}\:\mathrm{Atan}\:\mathrm{Btan}\:\mathrm{tan}\:\mathrm{C} \\ $$$$\mathrm{tan}\:\alpha\:=\mathrm{cot}\:\mathrm{Acot}\:\mathrm{B}\:\:\:\:\mathrm{tan}\:\beta\:=\mathrm{cot}\:\mathrm{Bcot}\:\mathrm{C}\:\:\:\:\mathrm{tan}\:\gamma\:=\mathrm{cot}\:\mathrm{Ccot}\:\mathrm{A} \\ $$$$\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta+\mathrm{tan}\:\gamma\:=\mathrm{1} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta+\gamma\right)\:=\frac{\Sigma\mathrm{tan}\:\alpha−\mathrm{tan}\:\alpha\mathrm{tan}\:\beta\mathrm{tan}\:\gamma}{\mathrm{1}−\Sigma\mathrm{tan}\:\alpha\mathrm{tan}\:\beta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\:\frac{\mathrm{1}−\left(\mathrm{cot}\:\mathrm{Acot}\:\mathrm{Bcot}\:\mathrm{C}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cot}\:\mathrm{Acot}\:\mathrm{Bcot}\:\mathrm{C}\left(\mathrm{cot}\:\mathrm{A}+\mathrm{cot}\:\mathrm{B}+\mathrm{cot}\:\mathrm{C}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{tan}\:\mathrm{Atan}\:\mathrm{Btan}\:\mathrm{C}−\mathrm{cot}\:\mathrm{Acot}\:\mathrm{Bcot}\:\mathrm{C}}{\Sigma\mathrm{tan}\:\mathrm{A}−\Sigma\mathrm{cot}\:\mathrm{A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{tan}\:\mathrm{Atan}\:\mathrm{Btan}\:\mathrm{C}−\mathrm{cot}\:\mathrm{Acot}\:\mathrm{Bcot}\:\mathrm{C}}{\Sigma\left(\frac{−\mathrm{cos}\:\mathrm{2A}}{\mathrm{sin}\:\mathrm{2A}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{8}\left(\mathrm{sin}\:\mathrm{Asin}\:\mathrm{Bsin}\:\mathrm{C}\right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{cos}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}\right)^{\mathrm{2}} }{−\Sigma\mathrm{sin}\:\mathrm{2A}\left(\mathrm{sin}\:\mathrm{2Bcos}\:\mathrm{2C}+\mathrm{sin}\:\mathrm{2Ccos}\:\mathrm{2B}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\Sigma\mathrm{sin}\:\mathrm{2A}\right)^{\mathrm{2}} −\left[\mathrm{1}+\Sigma\mathrm{cos}\:\mathrm{2A}\right]^{\mathrm{2}} \right\}}{\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}\:−\mathrm{4}\Sigma\mathrm{cos}\:\mathrm{2A}−\mathrm{4}\right\}}{\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}+\mathrm{8cos}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}{\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\:\frac{\mathrm{8cos}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}{\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}} \\ $$$$\alpha+\beta+\gamma\:=\mathrm{tan}^{−\mathrm{1}} \left\{\mathrm{1}+\frac{\mathrm{8cos}\:\mathrm{Acos}\:\mathrm{Bcos}\:\mathrm{C}}{\Sigma\mathrm{sin}\:^{\mathrm{2}} \mathrm{2A}}\right\}\:\:\mathrm{ans}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com