Question-192917

Question Number 192917 by BaliramKumar last updated on 31/May/23

Answered by AST last updated on 31/May/23

Let tanθ=t,cotθ=c; tc=1 t^2 +c^2 =(t+c)^2 −2tc=(t+c)^2 −2 t^5 +c^5 =(t+c)^5 −5t^4 c−10t^3 c^2 −10t^2 c^3 −5tc^4 ⇒2525=(t+c)^5 −5t^3 −10t−10c−5c^3 =(t+c)^5 −5(t+c)^3 +5(t+c)=2525 t+c=5 works⇒t^2 +c^2 =25−2=23

$${Let}\:{tan}\theta={t},{cot}\theta={c};\:{tc}=\mathrm{1} \\ $$$${t}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({t}+{c}\right)^{\mathrm{2}} −\mathrm{2}{tc}=\left({t}+{c}\right)^{\mathrm{2}} −\mathrm{2} \\ $$$${t}^{\mathrm{5}} +{c}^{\mathrm{5}} =\left({t}+{c}\right)^{\mathrm{5}} −\mathrm{5}{t}^{\mathrm{4}} {c}−\mathrm{10}{t}^{\mathrm{3}} {c}^{\mathrm{2}} −\mathrm{10}{t}^{\mathrm{2}} {c}^{\mathrm{3}} −\mathrm{5}{tc}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{2525}=\left({t}+{c}\right)^{\mathrm{5}} −\mathrm{5}{t}^{\mathrm{3}} −\mathrm{10}{t}−\mathrm{10}{c}−\mathrm{5}{c}^{\mathrm{3}} \\ $$$$=\left({t}+{c}\right)^{\mathrm{5}} −\mathrm{5}\left({t}+{c}\right)^{\mathrm{3}} +\mathrm{5}\left({t}+{c}\right)=\mathrm{2525} \\ $$$${t}+{c}=\mathrm{5}\:{works}\Rightarrow{t}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{25}−\mathrm{2}=\mathrm{23} \\ $$

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Question-192917

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