a-1-b-tan59-b-1-c-tan60-c-1-a-tan61-abc-2022-1-abc-2022-

Question Number 179195 by mathlove last updated on 26/Oct/22

$${a}+\frac{\mathrm{1}}{{b}}={tan}\mathrm{59} \\ $$$${b}+\frac{\mathrm{1}}{{c}}={tan}\mathrm{60} \\ $$$${c}+\frac{\mathrm{1}}{{a}}={tan}\mathrm{61} \\ $$$$\left({abc}\right)^{\mathrm{2022}} +\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{2022}} }=? \\ $$

Answered by Rasheed.Sindhi last updated on 26/Oct/22

a+(1/b)=tan59∧b+(1/c)=tan60∧c+(1/a)=tan61 (a+(1/b))(b+(1/c))(c+(1/a))=tan59tan60tan61 a+(1/b)+b+(1/c)+c+(1/a)+abc+(1/(abc))=tan59tan60tan61 tan59+tan60+tan61+abc+(1/(abc))=tan59tan60tan61 abc+(1/(abc)) =tan59tan60tan61−(tan59+tan60+tan61) =(√3) tan59tan61−(tan59+(√3) +tan61) =(√3) ( tan59tan61−1)−(tan59 +tan61) ....

$${a}+\frac{\mathrm{1}}{{b}}={tan}\mathrm{59}\wedge{b}+\frac{\mathrm{1}}{{c}}={tan}\mathrm{60}\wedge{c}+\frac{\mathrm{1}}{{a}}={tan}\mathrm{61} \\ $$$$\left({a}+\frac{\mathrm{1}}{{b}}\right)\left({b}+\frac{\mathrm{1}}{{c}}\right)\left({c}+\frac{\mathrm{1}}{{a}}\right)={tan}\mathrm{59}{tan}\mathrm{60}{tan}\mathrm{61} \\ $$$${a}+\frac{\mathrm{1}}{{b}}+{b}+\frac{\mathrm{1}}{{c}}+{c}+\frac{\mathrm{1}}{{a}}+{abc}+\frac{\mathrm{1}}{{abc}}={tan}\mathrm{59}{tan}\mathrm{60}{tan}\mathrm{61} \\ $$$${tan}\mathrm{59}+{tan}\mathrm{60}+{tan}\mathrm{61}+{abc}+\frac{\mathrm{1}}{{abc}}={tan}\mathrm{59}{tan}\mathrm{60}{tan}\mathrm{61} \\ $$$${abc}+\frac{\mathrm{1}}{{abc}} \\ $$$$\:\:={tan}\mathrm{59}{tan}\mathrm{60}{tan}\mathrm{61}−\left({tan}\mathrm{59}+{tan}\mathrm{60}+{tan}\mathrm{61}\right) \\ $$$$\:\:=\sqrt{\mathrm{3}}\:{tan}\mathrm{59}{tan}\mathrm{61}−\left({tan}\mathrm{59}+\sqrt{\mathrm{3}}\:+{tan}\mathrm{61}\right) \\ $$$$\:\:=\sqrt{\mathrm{3}}\:\left(\:{tan}\mathrm{59}{tan}\mathrm{61}−\mathrm{1}\right)−\left({tan}\mathrm{59}\:+{tan}\mathrm{61}\right) \\ $$$$…. \\ $$

Commented by som(math1967) last updated on 26/Oct/22

59+61=180−60 tan(61+59)=tan(180−60) ((tan61+tan59)/(1−tan61tan50))=−tan60 tan61+tan59+tan60=tan61tan60tan59 tan61+tan59+tan60−tan61tan60tan 59 =0 ∴ abc+(1/(abc))=0 ⇒ abc=−(1/(abc)) (abc)^2 =−1 (abc)^(2×1011) =(−1)^(1011) =−1 (abc)^(2022) +(1/((abc)^(2022) ))=−2

$$\:\mathrm{59}+\mathrm{61}=\mathrm{180}−\mathrm{60} \\ $$$${tan}\left(\mathrm{61}+\mathrm{59}\right)={tan}\left(\mathrm{180}−\mathrm{60}\right) \\ $$$$\frac{{tan}\mathrm{61}+{tan}\mathrm{59}}{\mathrm{1}−{tan}\mathrm{61}{tan}\mathrm{50}}=−{tan}\mathrm{60} \\ $$$${tan}\mathrm{61}+{tan}\mathrm{59}+{tan}\mathrm{60}={tan}\mathrm{61}{tan}\mathrm{60}{tan}\mathrm{59} \\ $$$${tan}\mathrm{61}+{tan}\mathrm{59}+{tan}\mathrm{60}−{tan}\mathrm{61}{tan}\mathrm{60tan}\:\mathrm{59} \\ $$$$\:=\mathrm{0} \\ $$$$\therefore\:{abc}+\frac{\mathrm{1}}{{abc}}=\mathrm{0} \\ $$$$\Rightarrow\:{abc}=−\frac{\mathrm{1}}{{abc}} \\ $$$$\left({abc}\right)^{\mathrm{2}} =−\mathrm{1} \\ $$$$\left({abc}\right)^{\mathrm{2}×\mathrm{1011}} =\left(−\mathrm{1}\right)^{\mathrm{1011}} =−\mathrm{1} \\ $$$$\:\left({abc}\right)^{\mathrm{2022}} +\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{2022}} }=−\mathrm{2} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Oct/22

$$\cap\boldsymbol{\mathrm{i}}\subset\in! \\ $$$$\top\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}! \\ $$

Commented by Sheshdevsahu last updated on 26/Oct/22

⇒(abc)^(2 ) = −1 ⇒(abc)^4 = 1 ⇒(abc)^(2022) +(1/((abc)^(2022) )) = {(abc)^4 }^(505) .(abc)^2 + (1/({(abc)}^(505) .(abc)^2 )) ⇒ = (1)^(505) .(−1) + (1/((1)^(505) (−1))) ⇒ = 1.(−1) + (1/(1.(−1))) ⇒ = −1 −1 ⇒ = −2 (ans)

$$ \\ $$$$\Rightarrow\left({abc}\right)^{\mathrm{2}\:} =\:−\mathrm{1} \\ $$$$\Rightarrow\left({abc}\right)^{\mathrm{4}} \:=\:\mathrm{1} \\ $$$$\Rightarrow\left({abc}\right)^{\mathrm{2022}} +\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{2022}} }\:=\:\left\{\left({abc}\right)^{\mathrm{4}} \right\}^{\mathrm{505}} .\left({abc}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\left\{\left({abc}\right)\right\}^{\mathrm{505}} .\left({abc}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:=\:\left(\mathrm{1}\right)^{\mathrm{505}} .\left(−\mathrm{1}\right)\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}\right)^{\mathrm{505}} \:\left(−\mathrm{1}\right)} \\ $$$$\Rightarrow\:\:=\:\mathrm{1}.\left(−\mathrm{1}\right)\:+\:\frac{\mathrm{1}}{\mathrm{1}.\left(−\mathrm{1}\right)} \\ $$$$\Rightarrow\:\:=\:−\mathrm{1}\:−\mathrm{1} \\ $$$$\Rightarrow\:\:=\:−\mathrm{2}\:\:\:\:\left({ans}\right) \\ $$

Commented by mathlove last updated on 26/Oct/22