Plz solve question 23 with explanation

sin4θa+cos4θb=1a+bsin4θa+cos2θ2b=1a+bsin4θa+1-sin2θ2b=1a+bsin4θa+1+sin4θ-2 sin2θb=1a+bb sin4θ+a+a sin4θ-2a sin2θab=1a+ba+b sin4θ-2a sin2θ+a=aba+ba+b2 sin4θ-2a+b a sin2θ+aa+b=aba+b sin2θ2-2 a a+b sin2θ+a2=0a+b sin2θ-a2=0a+b sin2θ-a=0sin2θ=aa+bTaking 4th power on both sides , we getsin8θ=aa+b4On dividing both sides by a3,we getsin8θa3=aa+b4   ...1Now,sin4θa+cos4θb=1a+bsin2θ2a+cos4θb=1a+b1-cos2θ2a+cos4θb=1a+b1+cos4θ-2 cos2θa+cos4θb=1a+bb+b cos4θ-2b cos2θ+a cos4θab=1a+ba+b cos4θ-2b cos2θ+b=aba+ba+b2 cos4θ-2a+b b cos2θ+ba+b=aba+b cos2θ2-2 b a+b cos2θ+b2=0a+b cos2θ-b2=0a+b cos2θ-b=0cos2θ=ba+bTaking 4th power on both sides , we getcos8θ=ba+b4 On dividing both sides by b3,we getcos8θb3=ba+b4    ...2Adding 1 and 2 , we getsin8θa3+cos8θb3=aa+b4+ba+b4sin8θa3+cos8θb3=a+ba+b4sin8θa3+cos8θb3=1a+b3Hence Proved.

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