Question

If $\frac{{\sin }^{4}A}{a}+\frac{{\cos }^{4}A}{b}=\frac{1}{a+b}$, then the value of $\frac{{\sin }^{8}A}{a^3}+\frac{{\cos }^{8}A}{b^3}$ is equal 5o

• A. $\frac{1}{(a+b{)}^3}$
• B. $\frac{a^3{b}^3}{(a+b{)}^3}$
• C. $\frac{a^2{b}^2}{(a+b{)}^2}$
• D. None of these

Answer 1

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

$\frac{\sin {}^{4}A}{a}+\frac{\cos {}^{4}A}{b}=\frac{1}{a+b}$

$\frac{\sin {}^{4}A}{a}+\frac{{(1-\sin {}^{2}A)}^2}{b}=\frac{1}{a+b}$

$\frac{\sin {}^{4}A}{a}+\frac{1+\sin {}^{4}A-2\sin {}^{2}A}{b}=\frac{1}{a+b}$

$b\sin {}^{4}A+a+a\sin {}^{4}A-2a\sin {}^{2}A=\frac{ab}{a+b}$

$(a+b)\sin {}^{4}A-2a\sin {}^{2}A+a-\frac{ab}{(a+b)}=0$

${(a+b)}^2\sin {}^{4}A-2a(a+b)\sin {}^{2}A+a(a+b)-ab=0$

${\left[(a+b)\sin {}^{2}A-a\right]}^2=0$

$(a+b)\sin {}^{2}A=a$

$\sin {}^{2}A=\frac{a}{a+b},\cos {}^{2}A=\frac{b}{a+b}$

$\frac{\sin {}^{8}A}{a^3}+\frac{\cos {}^{8}A}{b^3}=\frac{a^4}{{(a+b)}^4\times a^3}+\frac{b^4}{{(a+b)}^4\times b^3}=\frac{1}{{(a+b)}^3}$

Hence, $\frac{1}{(a+b{)}^3}$ is the correct answer.