If axcosθ+bysinθ=(a2−b2), and axsinθcos2θ−bycosθsin2θ=0, then (ax)2/3+(by)2/3=

From the second relation, we get sin2θsinθcos2θcosθ=byax;tan3θ=by/ax∴tanθ=(by)1/3/(ax)1/3∴sinθ=(by)1/3(ax)2/3+(by)2/31/2cosθ=(ax)1/3(ax)2/3+(by)2/31/2Putting for sinθ and cosθin axcosθ+bysinθ=a2−b2