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Question Number 143320 by mathdanisur last updated on 12/Jun/21
prove that: tan^2 36° + tan^2 72° = 5
$${prove}\:{that}:\:\:{tan}^{\mathrm{2}} \mathrm{36}°\:+\:{tan}^{\mathrm{2}} \mathrm{72}°\:=\:\mathrm{5} \\ $$
Answered by Ar Brandon last updated on 12/Jun/21
tan^2 36°+tan^2 72°=tan^2 36°+cot^2 18° =(((√(10−2(√5)))/( (√5)+1)))^2 +(((√(10+2(√5)))/( (√5)−1)))^2 =((10−2(√5))/(6+2(√5)))+((10+2(√5))/(6−2(√5))) =(((80−32(√5))+(80+32(√5)))/(16)) =((160)/(16))=10
$$\mathrm{tan}^{\mathrm{2}} \mathrm{36}°+\mathrm{tan}^{\mathrm{2}} \mathrm{72}°=\mathrm{tan}^{\mathrm{2}} \mathrm{36}°+\mathrm{cot}^{\mathrm{2}} \mathrm{18}° \\ $$$$=\left(\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}+\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\left(\mathrm{80}−\mathrm{32}\sqrt{\mathrm{5}}\right)+\left(\mathrm{80}+\mathrm{32}\sqrt{\mathrm{5}}\right)}{\mathrm{16}} \\ $$$$=\frac{\mathrm{160}}{\mathrm{16}}=\mathrm{10} \\ $$
Commented by mathdanisur last updated on 13/Jun/21
thanks Sir..
$${thanks}\:{Sir}.. \\ $$
Answered by MJS_new last updated on 13/Jun/21
it′s wrong
$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong} \\ $$
Answered by MJS_new last updated on 13/Jun/21
tan^2 x +tan^2 2x =10 tan x =t ((t^2 (t^4 −2t^2 +5))/((t^2 +1)^2 ))=10 t^6 −12t^4 +25t^2 −10=0 (t^2 −2)(t^4 −10t^2 +5)=0 t=±(√2)∨t=±(√(5−2(√5)))∨t=±(√(5+2(√5))) t=tan x ⇔ x=nπ+arctan t for 0≤x<90° we get x=36°∨x≈54.74°∨x=72°
$$\mathrm{tan}^{\mathrm{2}} \:{x}\:+\mathrm{tan}^{\mathrm{2}} \:\mathrm{2}{x}\:=\mathrm{10} \\ $$$$\mathrm{tan}\:{x}\:={t} \\ $$$$\frac{{t}^{\mathrm{2}} \left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{5}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{10} \\ $$$${t}^{\mathrm{6}} −\mathrm{12}{t}^{\mathrm{4}} +\mathrm{25}{t}^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{2}\right)\left({t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{5}\right)=\mathrm{0} \\ $$$${t}=\pm\sqrt{\mathrm{2}}\vee{t}=\pm\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}\vee{t}=\pm\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$${t}=\mathrm{tan}\:{x}\:\Leftrightarrow\:{x}={n}\pi+\mathrm{arctan}\:{t} \\ $$$$\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{90}°\:\mathrm{we}\:\mathrm{get} \\ $$$${x}=\mathrm{36}°\vee{x}\approx\mathrm{54}.\mathrm{74}°\vee{x}=\mathrm{72}° \\ $$
Commented by mathdanisur last updated on 13/Jun/21
Sir that is it cannot be 5.?
$${Sir}\:{that}\:{is}\:{it}\:{cannot}\:{be}\:\mathrm{5}.? \\ $$
Commented by MJS_new last updated on 13/Jun/21
no. just type it in any calculator.
$$\mathrm{no}.\:\mathrm{just}\:\mathrm{type}\:\mathrm{it}\:\mathrm{in}\:\mathrm{any}\:\mathrm{calculator}. \\ $$

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