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Q.

If sin4θ + cos4θ= l+3+ k sin2θ cos2θ then

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a

l + k = 0

b

l – k = 0

c

l + 2k = 0

d

2l + k = 0

answer is B.

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Detailed Solution

sin4θ+cos4θ=l+3+k sin2θ cos2θ sin2θ+cos2θ2-2sin2θ cos2θ =l+3+k sin2θ cos2θ 1-2sin2θ cos2θ=l+3+k sin2θ cos2θ l+3=1  l=-2 k=-2 l-k=0

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