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Question

If sin4 Aa+cos4 Ab=1a+b, then the value of sin8 Aa3+cos8 Ab3 is equal to


A

1(a+b)3

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B

a3b3(a+b)3

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C

a2b2(a+b)2

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D

None of these

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Solution

The correct option is A

1(a+b)3


It is given that sin4 Aa+cos4 Ab=1a+b

(1cos 2 A)24a+(1+cos 2 A)24b=1a+bb(a+b)(12 cos 2A+cos2 2A)+a(a+b)(1+2 cos 2A+cos2 2A)=4ab{b(a+b)+a(a+b)}cos2 2A+2(a+b)(ab)cos 2A +a(a+b)+b(a+b)4ab=0(a+b)2 cos2 2A+2(a+b)(ab)cos 2A+(ab)2=0{(a+b)cos 2A+(ab)}2=0 or cos 2A=bab+a

sin4 Aa+cos4 Ab=1a+b1a=1a+b b=0
Hence, sin8 Aa3+cos8 Ab3=(1cos 2A)416 a3+(1+cos 2A)416 b3=116a3[1bab+a]4+116b3[1+bab+a]4
=16a416a3(b+a)4+16b416b3(b+a)4=1(b+a)4(a+b)=1(a+b)3
Trick:Put A=90, then

sin8 Aa3+cos8 Ab3=1a3 which is given by option (a) Note : Students can check this question for other values of A also.


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