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Question Number 34647 by JOHNMASANJA last updated on 09/May/18

$$if{xsin}^3\theta +{ycos}^3\theta =sin\theta cos\theta$$$$andxsin\theta -ycos\theta =0$$$$provethat{x}^2+y^2=1$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/May/18

$$xsin\theta =ycos\theta$$$$\frac{x}{cos\theta }=\frac{y}{sin\theta }=k\left(assumed\right)$$$$x=kcos\theta ,y=ksin\theta$$$$puttinginfirstequation$$$$kcos\theta {sin}^3\theta +ksin\theta {cos}^3\theta =sin\theta cos\theta$$$$ksin\theta cos\theta ({sin}^2\theta +{cos}^2\theta )=sin\theta cos\theta$$$$sok=1$$$$hence{x}^2+y^2$$$$={cos}^2\theta +{sin}^2\theta$$$$=1$$

Answered by ajfour last updated on 09/May/18

$$x\sin \theta =y\cos \theta$$$$\Rightarrow x\sin \theta \left(\sin {}^2\theta \right)+y\cos \theta \left(\cos {}^2\theta \right)$$$$=\frac{x\sin \theta \cos \theta }{x}$$$$orx\sin {}^2\theta +x\cos {}^2\theta =\cos \theta$$$$\Rightarrow x=\cos \theta$$$$similarly$$$$y\sin {}^2\theta +y\cos {}^2\theta =\sin \theta$$$$\Rightarrow y=\sin \theta$$$$hence{\mathit{x}}^2+{\mathit{y}}^2=1.$$

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