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Question-70162

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Question Number 70162 by TawaTawa last updated on 01/Oct/19
Answered by mind is power last updated on 01/Oct/19
⇒1=(a+b)(((sin^4 (θ))/a)+((cos^4 (θ))/b))...S ⇒(b/a)Sin^4 (θ)+(a/b)cos^4 (θ)+cos^4 (θ)+sin^4 (θ)=1 t=(a/b) ⇒tsin^4 (θ)+((cos^4 (θ))/t)=1−(cos^4 (𝛉)+sin^4 (θ)) ⇒tsin^4 (θ)+((cos^4 (θ))/t)=2cos^2 θ)sin^2 (θ) ⇒t ((sin^2 (θ))/(cos^2 (θ)))+((cos^2 (θ))/(tsin^2 (θ)))=2 ⇒t.tan^2 (θ)+(1/(t tan^2 (θ)))=2 let u=t tan^2 (θ) ⇒u+(1/u)=2 ⇒((√u) −(1/( (√u))))^2 =0⇒u=1 t .tan^2 (θ)=1⇒(a/b)=((cos^2 (θ))/(sin^2 (θ)))=(1/(sin^2 (θ)))−1 ⇒Sin^2 (θ)=(b/(a+b)) cos^2 (θ)=(a/(a+b)) Sin^8 (θ)=(b^4 /((a+b)^4 )) cos^8 (θ)=(a^4 /((a+b)^3 )) ((Sin^8 (θ))/a^3 )+((cos^8 (θ))/b^3 )=(b^4 /(a^3 (a+b)^4 ))+(a^4 /(b^3 (a+b)^4 ))=(1/((a+b)^4 ))((b^4 /a^3 )+(a^4 /b^3 ))..H ((b^3 /a^3 )+(a^3 /b^3 ))^ =((b/a)+(a/b))^3 −3((a/b)+(b/a))=8−6=2 (a^2 /b^2 )+(b^2 /a^2 )=((a/b)+(b/a))^2 −2=2 (b^4 /a^3 )+(a^4 /b^3 )=(a+b)((b^3 /a^3 )+(a^3 /b^3 ) −(b^2 /a^2 )−(a^2 /b^2 ) +(b/a)+(a/b)−1)=(a+b)[2−2+2−1)=(a+b) H=(1/((a+b)^4 )).(a+b)=(1/((a+b)^3 ))
$$\Rightarrow\mathrm{1}=\left({a}+{b}\right)\left(\frac{{sin}^{\mathrm{4}} \left(\theta\right)}{{a}}+\frac{{cos}^{\mathrm{4}} \left(\theta\right)}{{b}}\right)…{S} \\ $$$$\Rightarrow\frac{{b}}{{a}}{Sin}^{\mathrm{4}} \left(\theta\right)+\frac{{a}}{{b}}{cos}^{\mathrm{4}} \left(\theta\right)+{cos}^{\mathrm{4}} \left(\theta\right)+{sin}^{\mathrm{4}} \left(\theta\right)=\mathrm{1} \\ $$$${t}=\frac{{a}}{{b}} \\ $$$$\Rightarrow{tsin}^{\mathrm{4}} \left(\theta\right)+\frac{{cos}^{\mathrm{4}} \left(\theta\right)}{{t}}=\mathrm{1}−\left(\boldsymbol{{cos}}^{\mathrm{4}} \left(\boldsymbol{\theta}\right)+{sin}^{\mathrm{4}} \left(\theta\right)\right) \\ $$$$\left.\Rightarrow{tsin}^{\mathrm{4}} \left(\theta\right)+\frac{{cos}^{\mathrm{4}} \left(\theta\right)}{{t}}=\mathrm{2}{cos}^{\mathrm{2}} \theta\right){sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\Rightarrow{t}\:\frac{{sin}^{\mathrm{2}} \left(\theta\right)}{{cos}^{\mathrm{2}} \left(\theta\right)}+\frac{{cos}^{\mathrm{2}} \left(\theta\right)}{{tsin}^{\mathrm{2}} \left(\theta\right)}=\mathrm{2} \\ $$$$\Rightarrow{t}.{tan}^{\mathrm{2}} \left(\theta\right)+\frac{\mathrm{1}}{{t}\:{tan}^{\mathrm{2}} \left(\theta\right)}=\mathrm{2} \\ $$$${let}\:{u}={t}\:{tan}^{\mathrm{2}} \left(\theta\right) \\ $$$$\Rightarrow{u}+\frac{\mathrm{1}}{{u}}=\mathrm{2}\:\:\Rightarrow\left(\sqrt{{u}}\:−\frac{\mathrm{1}}{\:\sqrt{{u}}}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{u}=\mathrm{1} \\ $$$${t}\:.{tan}^{\mathrm{2}} \left(\theta\right)=\mathrm{1}\Rightarrow\frac{{a}}{{b}}=\frac{{cos}^{\mathrm{2}} \left(\theta\right)}{{sin}^{\mathrm{2}} \left(\theta\right)}=\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\theta\right)}−\mathrm{1} \\ $$$$\Rightarrow{Sin}^{\mathrm{2}} \left(\theta\right)=\frac{{b}}{{a}+{b}} \\ $$$${cos}^{\mathrm{2}} \left(\theta\right)=\frac{{a}}{{a}+{b}} \\ $$$${Sin}^{\mathrm{8}} \left(\theta\right)=\frac{{b}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{4}} } \\ $$$${cos}^{\mathrm{8}} \left(\theta\right)=\frac{{a}^{\mathrm{4}} }{\left({a}+{b}\right)^{\mathrm{3}} } \\ $$$$\frac{{Sin}^{\mathrm{8}} \left(\theta\right)}{{a}^{\mathrm{3}} }+\frac{{cos}^{\mathrm{8}} \left(\theta\right)}{{b}^{\mathrm{3}} }=\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{3}} \left({a}+{b}\right)^{\mathrm{4}} }+\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{3}} \left({a}+{b}\right)^{\mathrm{4}} }=\frac{\mathrm{1}}{\left({a}+{b}\right)^{\mathrm{4}} }\left(\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{3}} }+\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{3}} }\right)..{H} \\ $$$$\left(\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }+\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }\right)^{} =\left(\frac{{b}}{{a}}+\frac{{a}}{{b}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)=\mathrm{8}−\mathrm{6}=\mathrm{2} \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\left(\frac{{a}}{{b}}+\frac{{b}}{{a}}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2} \\ $$$$\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{3}} }+\frac{{a}^{\mathrm{4}} }{{b}^{\mathrm{3}} }=\left({a}+{b}\right)\left(\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }+\frac{{a}^{\mathrm{3}} }{{b}^{\mathrm{3}} }\:−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:+\frac{{b}}{{a}}+\frac{{a}}{{b}}−\mathrm{1}\right)=\left({a}+{b}\right)\left[\mathrm{2}−\mathrm{2}+\mathrm{2}−\mathrm{1}\right)=\left({a}+{b}\right) \\ $$$${H}=\frac{\mathrm{1}}{\left({a}+{b}\right)^{\mathrm{4}} }.\left({a}+{b}\right)=\frac{\mathrm{1}}{\left({a}+{b}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TawaTawa last updated on 01/Oct/19
Wow, God bless you sir
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mind is power last updated on 01/Oct/19
y′re Welcom
$${y}'{re}\:{Welcom} \\ $$$$ \\ $$

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