If sin4Aa+cos4Ab=1a+b,then sin8Aa3+cos8Ab3=

sin4Aa+cos4Ab=1a+b⇒ 1a1−cos2A22+1b1+cos 2A22+1b1+cos2A22=1a+b⇒b(a+b)(1−cos2A)2+a(a+b)(1+cos2A)2=4ab⇒[b(a+b)+a(a+b)]cos22A+[a(a+b)−b(a+b)]2cos 2A+b(a+b)+a(a+b)−4ab=0⇒(a+b)2cos22A+2+2(a2−b2)cos2A+(a−b)2=0⇒[(a+b)cos 2A+(a−b)]2=0⇒(a+b)cos2A+(a−b)=0⇒cos 2A=b−ab+a.sin8Aa3+cos8Ab3=1a31−cos2A24+1b31+cos2A24=116a31−b−ab+a+116b31+b−ab+a4=116a32ab+a4+116b32bb+a4=16a416a3(a+b)4+16b416b3(a+b)4=a(a+b)4+b(a+b)4=1(a+b)3.

If sin4Aa+cos4Ab=1a+b,then sin8Aa3+cos8Ab3=

sin4Aa+cos4Ab=1a+b⇒ 1a1−cos2A22+1b1+cos 2A22+1b1+cos2A22=1a+b⇒b(a+b)(1−cos2A)2+a(a+b)(1+cos2A)2=4ab⇒[b(a+b)+a(a+b)]cos22A+[a(a+b)−b(a+b)]2cos 2A+b(a+b)+a(a+b)−4ab=0⇒(a+b)2cos22A+2+2(a2−b2)cos2A+(a−b)2=0⇒[(a+b)cos 2A+(a−b)]2=0⇒(a+b)cos2A+(a−b)=0⇒cos 2A=b−ab+a.sin8Aa3+cos8Ab3=1a31−cos2A24+1b31+cos2A24=116a31−b−ab+a+116b31+b−ab+a4=116a32ab+a4+116b32bb+a4=16a416a3(a+b)4+16b416b3(a+b)4=a(a+b)4+b(a+b)4=1(a+b)3.

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$I f \frac{\sin^{4} A}{a} + \frac{\cos^{4} A}{b} = \frac{1}{a + b}, t h e n \frac{\sin^{8} A}{a^{3}} + \frac{\cos^{8} A}{b^{3}} =$

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$\frac{1}{{(a + b)}^{2}}$

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

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

$\frac{1}{{(a - b)}^{2}}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{\sin^{4} A}{a} + \frac{\cos^{4} A}{b} = \frac{1}{a + b} \Rightarrow \frac{1}{a} {(\frac{1 - \cos 2 A}{2})}^{2} + \frac{1}{b} {(\frac{1 + \cos 2 A}{2})}^{2} + \frac{1}{b} {(\frac{1 + \cos 2 A}{2})}^{2} = \frac{1}{a + b} \\ \Rightarrow b (a + b) {(1 - \cos 2 A)}^{2} + a (a + b) {(1 + \cos 2 A)}^{2} = 4 a b \\ \Rightarrow [b (a + b) + a (a + b)] \cos^{2} 2 A + [a (a + b) - b (a + b)] 2 \cos 2 A + b (a + b) + a (a + b) - 4 a b = 0 \\ \Rightarrow {(a + b)}^{2} \cos^{2} 2 A + 2 + 2 (a^{2} - b^{2}) \cos 2 A + {(a - b)}^{2} = 0 \Rightarrow {[(a + b) \cos 2 A + (a - b)]}^{2} = 0 \\ \Rightarrow (a + b) \cos 2 A + (a - b) = 0 \Rightarrow \cos 2 A = \frac{b - a}{b + a} . \\ \frac{\sin^{8} A}{a^{3}} + \frac{\cos^{8} A}{b^{3}} = \frac{1}{a^{3}} {(\frac{1 - \cos 2 A}{2})}^{4} + \frac{1}{b^{3}} {(\frac{1 + \cos 2 A}{2})}^{4} \\ = \frac{1}{16 a^{3}} (1 - \frac{b - a}{b + a}) + \frac{1}{16 b^{3}} {(1 + \frac{b - a}{b + a})}^{4} = \frac{1}{16 a^{3}} {(\frac{2 a}{b + a})}^{4} + \frac{1}{16 b^{3}} {(\frac{2 b}{b + a})}^{4} \\ = \frac{16 a^{4}}{16 a^{3} {(a + b)}^{4}} + \frac{16 b^{4}}{16 b^{3} {(a + b)}^{4}} = \frac{a}{{(a + b)}^{4}} + \frac{b}{{(a + b)}^{4}} = \frac{1}{{(a + b)}^{3}} . \end{array}$