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

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

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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 $\sin^2 A = \frac{1 - \cos 2A}{2}$ 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$$(a + b)^2 \cos^2 2A + 2(a + b)(a - b) \cos 2A + (a - b)^2 = 0$This is a perfect square: $((a + b) \cos 2A + (a - b))^2 = 0$So,$\cos 2A = \frac{b - a}{a + b}$Now,$\sin^2 A = \frac{1 - \cos 2A}{2} = \frac{1 - \frac{b - a}{a + b}}{2} = \frac{a}{a + b}$ and $\cos^2 A = \frac{1 + \cos 2A}{2} = \frac{1 + \frac{b - a}{a + b}}{2} = \frac{b}{a + b}$Therefore,$\frac{\sin^8 A}{a^3} + \frac{\cos^8 A}{b^3} = \frac{(\frac{a}{a + b})^4}{a^3} + \frac{(\frac{b}{a + b})^4}{b^3} = \frac{a^4}{a^3(a + b)^4} + \frac{b^4}{b^3(a + b)^4} = \frac{a + b}{(a + b)^4} = \frac{1}{(a + b)^3}$

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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