Question

If $\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{a+b}$, prove that
i) $\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}=\frac{1}{{(a+b)}^3}$
ii) $\frac{{\sin }^{4n}\theta }{a^{2n-1}}+\frac{{\cos }^{4n}\theta }{b^{2n-1}}=\frac{1}{{(a+b)}^{2n-1}},n\in N$

Answer 1

$\begin{array}{cc}\frac{{\sin }^4\theta }{a}+\frac{{\cos }^4\theta }{b}=\frac{1}{a+b}\to \left(i\right)& \\ provethat\left(i\right)\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}=\frac{1}{{\left(a+b\right)}^3}& \\ So,multiply\left(a+b\right)byequation\left(i\right),weget& \\ \left(a+b\right)\frac{{\sin }^4\theta }{a}+\left(a+b\right)\frac{{\cos }^4\theta }{b}=1& \\ \left(1+\frac{b}{a}\right){\sin }^4\theta +\left(\frac{a}{b}+1\right){\cos }^4\theta =1& \\ \Rightarrow \frac{b}{a}{\sin }^4\theta +\frac{a}{b}{\cos }^4\theta ={\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-\left({\sin }^4\theta +{\cos }^4\theta \right)& \left[\because 1={\cos }^2\theta +{\sin }^2\theta \right]\\ \Rightarrow \frac{b}{a}{\sin }^4\theta +\frac{a}{b}{\cos }^4\theta =2{\sin }^2\theta .{\cos }^2\theta & \\ \Rightarrow {\left(\frac{\sqrt{b}}{a}{\sin }^2\theta \right)}^2+{\left(\frac{\sqrt{a}}{b}{\cos }^2\theta \right)}^2-2\times \sqrt{\frac{b}{a}}\times \sqrt{\frac{a}{b}}{\sin }^2\theta .{\cos }^2\theta =0& \\ \left(\sqrt{\frac{b}{a}}{\sin }^2\theta -\sqrt{\frac{a}{b}}{\cos }^2\theta \right)=0& \\ \Rightarrow \sqrt{\frac{b}{a}}{\sin }^2\theta =\sqrt{\frac{a}{b}}{\cos }^2\theta & \\ \Rightarrow \frac{b}{a}{\sin }^4\theta =\frac{a}{b}{\cos }^4\theta & \\ \Rightarrow \frac{b^2}{a^2}{\sin }^4\theta ={\cos }^4\theta \Rightarrow \frac{{\sin }^2\theta }{a}=\frac{{\cos }^2\theta }{b}& \\ \frac{{\sin }^2\theta }{a}=\frac{{\cos }^2\theta }{b}=\frac{{\sin }^2\theta +{\cos }^2\theta }{a+b}\Rightarrow \frac{1}{a+b}& \\ So,\frac{{\sin }^8\theta }{a^3}+\frac{{\cos }^8\theta }{b^3}\Rightarrow {\sin }^2\theta .\frac{{\sin }^6\theta }{a^3}+{\cos }^2\theta .\frac{{\cos }^6\theta }{b^3}& \\ \Rightarrow {\sin }^2\theta {\left(\frac{{\sin }^2\theta }{a}\right)}^3+{\cos }^2\theta .{\left(\frac{{\cos }^2\theta }{b}\right)}^3\Rightarrow {\sin }^2\theta {\left(\frac{1}{a+b}\right)}^3+{\cos }^2\theta {\left(\frac{1}{a+b}\right)}^3& \\ \Rightarrow \frac{1}{{\left(a+b\right)}^3}\left[{\sin }^2\theta +{\cos }^2\theta \right]& \\ \frac{{\sin }^{8}a}{a^3}+\frac{{\cos }^8\theta }{a^3}\Rightarrow \frac{1}{{\left(a+b\right)}^3}& \end{array}$