Question

If $\frac{{\sin }^{4}A}{a}+\frac{{\cos }^{4}A}{b}=\frac{1}{(a+b)}$, then the value of $\left(\frac{{\sin }^{8}A}{a^3}\right)+\left(\frac{{\cos }^{8}A}{b^3}\right)$ is equal to

• A. $\frac{1}{{\left(a+b\right)}^3}$
• 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

$\frac{1}{{\left(a+b\right)}^3}$

Explanation for the correct option.

Step 1: Find the value of ${\sin }^{2}A$

Given that, $\frac{{\sin }^{4}A}{a}+\frac{{\cos }^{4}A}{b}=\frac{1}{(a+b)}$.

$\begin{array}{rcl}\frac{{\sin }^{4}A}{a}+\left[\frac{{\left(1-{\sin }^{2}A\right)}^2}{b}\right]& =& \frac{1}{(a+b)}\\ \frac{{\sin }^4\mathrm{A}}{\mathrm{a}}+\frac{\left[\left(1+{\sin }^4\mathrm{A}-2{\sin }^2\mathrm{A}\right)\right]}{b}& =& \frac{1}{(a+b)}\\ \frac{{\sin }^4\mathrm{A}}{\mathrm{a}}+\frac{1}{b}+\frac{{\sin }^{4}A}{b}-\frac{2{\sin }^{2}A}{b}& =& \frac{1}{(a+b)}\\ {\sin }^{4}A(\frac{1}{a}+\frac{1}{b})-\frac{2{\sin }^{2}A}{b}& =& \frac{1}{(a+b)}-\frac{1}{b}\\ \frac{\left[{\sin }^4\mathrm{A}(\mathrm{a}+\mathrm{b})\right]}{ab}-\frac{2{\mathrm{asin}}^2\mathrm{A}}{\mathrm{ab}}& =& \frac{(\mathrm{b}-\mathrm{a}-\mathrm{b})}{(\mathrm{a}+\mathrm{b})\mathrm{b}}\\ (a+b{)}^2{\left({\sin }^{2}A\right)}^2-2a(a+b){\sin }^{2}A& =& -a^2\\ {\left\{(a+b){\sin }^{2}A\right\}}^2-2a(a+b){\sin }^{2}A+a^2& =& 0\\ {\left\{(a+b){\sin }^{2}A-a\right\}}^2& =& 0\\ (a+b){\sin }^{2}A-a& =& 0\end{array}$

Step 2: Find the value of $\left(\frac{{\sin }^{8}A}{a^3}\right)+\left(\frac{{\cos }^{8}A}{b^3}\right)$

Now find the value of ${\cos }^{2}A$.

$\begin{array}{rcl}\therefore {\cos }^{2}A& =& 1-{\sin }^{2}A\\ & =& 1-\frac{{\displaystyle a}}{(a+b)}\\ & =& \frac{{\displaystyle b}}{(a+b)}\end{array}$

$\begin{array}{rcl}\therefore \frac{{\sin }^8\mathrm{A}}{{\mathrm{a}}^3}+\frac{{\cos }^8\mathrm{A}}{{\mathrm{b}}^3}& =& \frac{a}{(a+b{)}^4}+\frac{b}{(a+b{)}^4}\\ & =& \frac{a+b}{{\left(a+b\right)}^4}\\ & =& \frac{1}{{\left(a+b\right)}^3}\end{array}$

Hence, option A is correct.