Question

If $x\sin \theta -y\cos \theta =0$ and $x{\sin }^3\theta +y{\cos }^3\theta =\sin \theta \cos \theta ,$ then $x^2+y^2$ is equal to

• A. $0$
• B. $2$
• C. $4$
• D. $1$

Answer 1

Answer: (D)

Given, $x\sin \theta -y\cos \theta =0$ ..(i)
and $x{\sin }^3\theta +y{\cos }^3\theta =\sin \theta \cos \theta$ ..(ii)
On multiplying Eq. (i) by ${\sin }^2\theta$
and subtracting from Eq. (ii), we get
$y{\cos }^3\theta +y\cos {\sin }^2\theta =\sin \theta \cos \theta$
$\Rightarrow$ $y\cos \theta (1)=sin\theta cos\theta$
$\Rightarrow$ $y=\sin \theta$
From Eq. (i), $x\sin \theta =\sin \theta \cos \theta$
$\Rightarrow$ $x=\cos \theta$
Now, $x^2+y^2={\cos }^2\theta +{\sin }^2\theta =1$