Question

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

• A. $\frac{1}{{(a+b)}^3}$
• B. $\frac{1}{{(a-b)}^3}$
• C. $\frac{1}{{(a+b)}^2}$
• D. $\frac{1}{{(a-b)}^2}$

Answer 1

The correct option is A $\frac{1}{{(a+b)}^3}$

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

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

$\Rightarrow (a+b)cos2A+(a-b)=0\Rightarrow cos2A=\frac{b-a}{b+a}$

$\frac{si{n}^{8}A}{a^3}+\frac{co{s}^{8}A}{b^3}=\frac{1}{a^3}{\left(\frac{1-cos2A}{2}\right)}^4+\frac{1}{b^3}{\left(\frac{1+cos2A}{2}\right)}^4$

=$\frac{1}{16a^3}{(1-\frac{b-a}{b+a})}^4+\frac{1}{16b^3}{(1+\frac{b-a}{b+a})}^4=\frac{1}{16a^3}{\left(\frac{2a}{b+a}\right)}^4+\frac{1}{16b^3}{\left(\frac{2b}{b+a}\right)}^4$

=$\frac{16a^4}{16a^3(a+b{)}^4}+\frac{16b^4}{16b^3(a+b{)}^4}=\frac{a}{(a+b{)}^4}+\frac{b}{(a+b{)}^4}=\frac{1}{(a+b{)}^3}$