Question

$If\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b}=\frac{1}{a+b^{\prime }},provethat\frac{si{n}^8\theta }{a^8}+\frac{co{s}^8\theta }{b^8}=\frac{1}{a+b^8}$

Answer 1

We have

$\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b}=\frac{1}{a+b}\Rightarrow (a+b)(\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b})=1\Rightarrow (a+b)(\frac{si{n}^4\theta }{a}+\frac{co{s}^4\theta }{b})(si{n}^2\theta +co{s}^2\theta {)}^2\Rightarrow \left(\frac{a+b}{a}\right)si{n}^4\theta +(\frac{a}{b}+1)co{s}^4\theta =si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta \Rightarrow \frac{b}{a}si{n}^4\theta +\frac{a}{b}co{s}^4\theta -2si{n}^2\theta co{s}^2\theta =0\Rightarrow {(\sqrt{\frac{b}{a}}si{n}^2\theta -\sqrt{\frac{a}{b}}co{s}^2\theta )}^2=0$

$\Rightarrow {(\sqrt{\frac{b}{a}}si{n}^2\theta -\sqrt{\frac{a}{b}}co{s}^2\theta )}^2=0\Rightarrow \sqrt{\frac{b}{a}}si{n}^2\theta -\sqrt{\frac{a}{b}}co{s}^2\theta =0\Rightarrow \sqrt{\frac{b}{a}}si{n}^2\theta =\sqrt{\frac{a}{b}}co{s}^2\theta \Rightarrow \frac{si{n}^2\theta }{co{s}^2\theta }=\frac{a}{b}\Rightarrow \frac{si{n}^2\theta }{a}=\frac{co{s}^2\theta }{b}$

Using ratio and proportion,$\frac{si{n}^2\theta }{a}=\frac{co{s}^2\theta }{b}=\frac{si{n}^2\theta +co{s}^2\theta }{a+b}=\frac{1}{a+b}$
$\Rightarrow si{n}^2\theta =\frac{a}{a+b}andco{s}^2\theta =\frac{b}{a+b}So,\frac{si{n}^3\theta }{a^3}+\frac{co{s}^3\theta }{b^3}=(si{n}^2\theta {)}^4\frac{1}{a^3}+(co{s}^2\theta {)}^4\frac{1}{b^3}=\frac{1}{a^3}{\left(\frac{a}{a+b}\right)}^4+\frac{1}{b^3}{\left(\frac{b}{(a+b)}\right)}^4=\frac{1}{(a+b{)}^3}$