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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

$$\Rightarrow 1=(a+b)(\frac{{sin}^4\left(\theta \right)}{a}+\frac{{cos}^4\left(\theta \right)}{b})...S$$$$\Rightarrow \frac{b}{a}{Sin}^4\left(\theta \right)+\frac{a}{b}{cos}^4\left(\theta \right)+{cos}^4\left(\theta \right)+{sin}^4\left(\theta \right)=1$$$$t=\frac{a}{b}$$$$\Rightarrow {tsin}^4\left(\theta \right)+\frac{{cos}^4\left(\theta \right)}{t}=1-({\mathit{c}\mathit{o}\mathit{s}}^4\left(\mathit{\theta }\right)+{sin}^4\left(\theta \right))$$$$\Rightarrow {tsin}^4\left(\theta \right)+\frac{{cos}^4\left(\theta \right)}{t}=2{cos}^2\theta ){sin}^2\left(\theta \right)$$$$\Rightarrow t\frac{{sin}^2\left(\theta \right)}{{cos}^2\left(\theta \right)}+\frac{{cos}^2\left(\theta \right)}{{tsin}^2\left(\theta \right)}=2$$$$\Rightarrow t.{tan}^2\left(\theta \right)+\frac{1}{t{tan}^2\left(\theta \right)}=2$$$$letu=t{tan}^2\left(\theta \right)$$$$\Rightarrow u+\frac{1}{u}=2\Rightarrow {(\sqrt{u}-\frac{1}{\sqrt{u}})}^2=0\Rightarrow u=1$$$$t.{tan}^2\left(\theta \right)=1\Rightarrow \frac{a}{b}=\frac{{cos}^2\left(\theta \right)}{{sin}^2\left(\theta \right)}=\frac{1}{{sin}^2\left(\theta \right)}-1$$$$\Rightarrow {Sin}^2\left(\theta \right)=\frac{b}{a+b}$$$${cos}^2\left(\theta \right)=\frac{a}{a+b}$$$${Sin}^8\left(\theta \right)=\frac{b^4}{{(a+b)}^4}$$$${cos}^8\left(\theta \right)=\frac{a^4}{{(a+b)}^3}$$$$\frac{{Sin}^8\left(\theta \right)}{a^3}+\frac{{cos}^8\left(\theta \right)}{b^3}=\frac{b^4}{a^3{(a+b)}^4}+\frac{a^4}{b^3{(a+b)}^4}=\frac{1}{{(a+b)}^4}(\frac{b^4}{a^3}+\frac{a^4}{b^3})..H$$$${(\frac{b^3}{a^3}+\frac{a^3}{b^3})}^{}={(\frac{b}{a}+\frac{a}{b})}^3-3(\frac{a}{b}+\frac{b}{a})=8-6=2$$$$\frac{a^2}{b^2}+\frac{b^2}{a^2}={(\frac{a}{b}+\frac{b}{a})}^2-2=2$$$$\frac{b^4}{a^3}+\frac{a^4}{b^3}=(a+b)(\frac{b^3}{a^3}+\frac{a^3}{b^3}-\frac{b^2}{a^2}-\frac{a^2}{b^2}+\frac{b}{a}+\frac{a}{b}-1)=(a+b)[2-2+2-1)=(a+b)$$$$H=\frac{1}{{(a+b)}^4}.(a+b)=\frac{1}{{(a+b)}^3}$$$$$$$$$$$$$$$$$$$$$$

Commented by TawaTawa last updated on 01/Oct/19

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by mind is power last updated on 01/Oct/19

$$y^{\prime }reWelcom$$$$$$

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