Question

If
$\frac{si{n}^{4}A}{a}+\frac{co{s}^{4}A}{b}=\frac{1}{a+b},thenthevalueof\frac{si{n}^{8}A}{a^3}+\frac{co{s}^{8}A}{b^3}$
$isequalto$

Answer 1

It is given that
$\frac{si{n}^{4}A}{a}+\frac{co{s}^{4}A}{b}=\frac{1}{a+b}$

$\Rightarrow \frac{(1-cos2A{)}^2}{4a}+\frac{(1+cos2A{)}^2}{4b}=\frac{1}{a+b}\Rightarrow b(a+b)(1-2cos2A+co{s}^{2}2A)+a(a+b)(1+2cos2A+co{s}^{2}2A)=4ab\Rightarrow \left\{b\right(a+b)+a(a+b\left)\right\}co{s}^{2}2A+2(a+b)(a-b)cos2A+a(a+b)+b(a+b)-4ab=0\Rightarrow (a+b{)}^{2}co{s}^{2}2A+2(a+b\left)\right(a-b)cos2A+(a-b{)}^2=0\Rightarrow {\left\{\right(a+b)cos2A+(a-b\left)\right\}}^2=0\Rightarrow cos2A=\frac{b-a}{b+a}$

$Hence,\Rightarrow \frac{si{n}^{8}A}{a^3}+\frac{co{s}^{8}A}{b^3}\Rightarrow \frac{(1-cos2A{)}^4}{16a^3}+\frac{(1+cos2A{)}^4}{16b^3}\Rightarrow \frac{1}{16a^3}{[1-\frac{b-a}{b+a}]}^4+\frac{1}{16b^3}{[1+\frac{b-a}{b+a}]}^4$
$\Rightarrow \frac{16a^4}{16a^3(b+a{)}^4}+\frac{16b^4}{16b^3(b+a{)}^4}\Rightarrow \frac{1}{(b+a{)}^4}(a+b)\Rightarrow \frac{1}{(a+b{)}^3}$

$Trick:PutA={90}^{\circ },then$

$\therefore \frac{si{n}^{8}A}{a^3}+\frac{co{s}^{8}A}{b^3}=\frac{1}{a^3}$
which is given by option (a)
Note : Students can check this question for other values of A also.