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Question

If sin4θa+cos4θb=1a+b,then

A
cos2θ=ba+b
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B
sin2θ=aa+b
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C
cos6θ+sin6θ=a2ab+b2a2+2ab+b2
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D
cos8θb3+sin8θa3=1(a+b)3
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Solution

The correct options are
A cos2θ=ba+b
B sin2θ=aa+b
C cos6θ+sin6θ=a2ab+b2a2+2ab+b2
D cos8θb3+sin8θa3=1(a+b)3
Let sin2θ=t
sin4θa+cos4θb=1a+b
t2a+(1t)2b=1a+b
t2a+t22t+1b=1a+b

bt2+at22at+a=aba+b
(a+b)t22at+a2a+b=0

t=2a±4a24a22(a+b)
sin2θ=t=aa+b ....................(1)
cos2θ=ba+b ......................(2)
Now using (1) and (2) we can also deduce the result for option C and option D.

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