Question

If $\frac{si{n}^{4}A}{a}+\frac{co{s}^{4}A}{b}=\frac{1}{a+b},$ then the value of $\frac{si{n}^{8}A}{a^3}+\frac{co{s}^{8}A}{b^3}$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

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

$\Rightarrow \frac{(1-cos2A{)}^2}{4a}+\frac{(1+cos2A{)}^2}{4b}=\frac{1}{a+b}$

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

$\Rightarrow \left\{b\right(a+b)+a(a+b\left)\right\}co{s}^{2}2A+2(a+b)(a-b)cos2A$