If frac{{sin^4 A}}{a} + frac{{cos^4 A}}{b} = | vedclass

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(A) Given: $\frac{{\sin^4 A}}{a} + \frac{{\cos^4 A}}{b} = \frac{1}{{a + b}}$Using the identity $\sin^2 A = \frac{{1 - \cos 2A}}{2}$ and $\cos^2 A = \f

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

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(A) Given: $\frac{{\sin^4 A}}{a} + \frac{{\cos^4 A}}{b} = \frac{1}{{a + b}}$Using the identity $\sin^2 A = \frac{{1 - \cos 2A}}{2}$ and $\cos^2 A = \frac{{1 + \cos 2A}}{2}$,we have:$\frac{{(1 - \cos 2A)^2}}{{4a}} + \frac{{(1 + \cos 2A)^2}}{{4b}} = \frac{1}{{a + b}}$Multiplying by $4ab(a + b)$:$b(a + b)(1 - 2\cos 2A + \cos^2 2A) + a(a + b)(1 + 2\cos 2A + \cos^2 2A) = 4ab$Expanding and rearranging terms:$(a + b)^2 \cos^2 2A + 2(a + b)(a - b) \cos 2A + (a - b)^2 = 0$This is a perfect square:$((a + b) \cos 2A + (a - b))^2 = 0$Thus,$\cos 2A = \frac{{b - a}}{{a + b}}$.Now,calculate $\frac{{\sin^8 A}}{{a^3}} + \frac{{\cos^8 A}}{{b^3}}$:$= \frac{{(1 - \cos 2A)^4}}{{16a^3}} + \frac{{(1 + \cos 2A)^4}}{{16b^3}}$Substituting $\cos 2A = \frac{{b - a}}{{a + b}}$:$= \frac{1}{{16a^3}} \left( 1 - \frac{{b - a}}{{a + b}} \right)^4 + \frac{1}{{16b^3}} \left( 1 + \frac{{b - a}}{{a + b}} \right)^4$$= \frac{1}{{16a^3}} \left( \frac{{2a}}{{a + b}} \right)^4 + \frac{1}{{16b^3}} \left( \frac{{2b}}{{a + b}} \right)^4$$= \frac{{16a^4}}{{16a^3(a + b)^4}} + \frac{{16b^4}}{{16b^3(a + b)^4}} = \frac{a + b}{{(a + b)^4}} = \frac{1}{{(a + b)^3}}$.

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 to

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