Question

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

• A. $2$
• B. $0$
• C. $3$
• D. $4$
• E. $1$

Answer 1

The correct option is E

$1$

Explanation for the correct option.

$\begin{array}{rcl}x{\sin }^3\theta +y{\cos }^3\theta & =& \sin \theta \cos \theta \\ \Rightarrow x\left(\sin \theta \right){\sin }^2\theta +y\left(\cos \theta \right){\cos }^2\theta & =& \sin \theta \cos \theta \\ & \Rightarrow & x\left(\sin \theta \right){\sin }^2\theta +x\left(\sin \theta \right){\cos }^2\theta =\sin \theta \cos \theta \left[\mathrm{Given}x\sin \theta =y\cos \theta \right]\\ \Rightarrow x\sin \theta \left({\sin }^2\theta +{\cos }^2\theta \right)& =& \sin \theta \cos \theta \\ \Rightarrow x\sin \theta & =& \sin \theta \cos \theta \left[\because si{n}^2\theta +co{s}^2\theta =1\right]\\ \Rightarrow x& =& \cos \theta \end{array}$

Now, we have

$\begin{array}{rcl}x\sin \theta & =& y\cos \theta \\ & \Rightarrow & \cos \theta \sin \theta =y\cos \theta \left[\mathrm{By}\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}x\right]\\ \Rightarrow y& =& \sin \theta \end{array}$

So,

$\begin{array}{rcl}{x}^2+y^2& =& {\cos }^2\theta +{\sin }^2\theta \\ & =& 1\end{array}$

Hence, option E is correct.