Here:
a sin3 x + b cos3x=sin X cos X
a sin X=bcosX
From the second equation a sin X = b cos X, we get:
a = (b cos X)/ sinX
Substitute a = (b cos X) / sin X into the first equation;
((b cos X)/ sinX) sin3 x +b cos3 x= sinX cosX
On Simplifying:
b cos X sin2x +bcos3x=sinXcosX
Factor out b cos X:
b cosX(sin2X + cos2x) = sinX cosx
Since sin2 x + cos2x=1, the equation becomes;
b cos X= sin X cos X
Now, divide both sides by cos X(since cos X *0):
b=sin X
Substitute b = sin X into the equation a =(b cos X)/ sin X:
a =(sin X cos X) / sin X = cos X
On calculating a2+b2:
a2+b2=(cosX)2+(sin x)2=1
Option (b) is the correct answer.
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