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यदि $x{\sin ^3}\alpha + y{\cos ^3}\alpha = \sin \alpha \cos \alpha $ व $x\sin \alpha - y\cos \alpha = 0,$ तो ${x^2} + {y^2} = $
$-1$
$±1$
$1$
इनमें से कोई नहीं
Solution
यहाँ $x\,{\sin ^3}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $…..$(i)$
एवं $x\,\sin \,\alpha – y\,\cos \,\alpha = 0$…..$(ii)$
अब $(ii)$ से, $x\,\sin \,\alpha = y\,\cos \,\alpha $
यह मान $(i)$ में रखने पर,
$ \Rightarrow \,\,y\,\cos \alpha \,{\sin ^2}\alpha + y\,{\cos ^3}\alpha = \sin \,\alpha \,\cos \,\alpha $
$ \Rightarrow \,\,y\,\cos \alpha \,\left\{ {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right\} = \sin \,\alpha \,\cos \,\alpha $
$ \Rightarrow \,\,y\,\cos \,\alpha = \sin \,\alpha \,\cos \,\alpha \,$
$\Rightarrow \,\,y = \sin \,\alpha $ एवं $x = \cos \,\alpha $
अत:, ${x^2} + {y^2} = {\sin ^2}\alpha + {\cos ^2}\alpha = 1.$
