Question

If $x{\sin }^3\alpha +y{\cos }^3\alpha =\sin \alpha \cos \alpha$ and $x\sin \alpha =y\cos \alpha$, then $x^2+y^2=$

• A. $-1$
• B. $\pm 1$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

$=x\sin \alpha =y\cos \alpha$.........(1)

$=x^3{\sin }^3\alpha =y^3{\cos }^{3}d$

$x{\sin }^3\alpha +y{\cos }^3\alpha =\sin \alpha \cos \alpha$

$\left(x\sin \alpha \right)\left({\sin }^2\alpha \right)+y{\cos }^3\alpha =\sin \alpha \cos \alpha$

$\left(y\cos \alpha \right)\left({\sin }^2\alpha \right)+\left(y\cos \alpha \right)\left({\cos }^2\alpha \right)=\sin 2\alpha$ From (1)

$\left(y\cos \alpha \right)({\sin }^2\alpha +{\cos }^2\alpha )=\sin \alpha \cos \alpha$

$\Rightarrow y\cos \alpha =\sin \alpha \cos \alpha$

$\Rightarrow y=\sin \alpha$........(2)

put 2 in 1

$x\sin \alpha =\left(\sin \alpha \right)\cos \alpha$

$\Rightarrow x=\cos \alpha$

$x^2+y^2={\sin }^2\alpha +{\cos }^2\alpha =1$