Q42 / 125 · BITSAT 2013 · Mathematics
If $x \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$ and
$x \sin \theta=y \cos \theta,$ then $x^{2}+y^{2}=$
- Correct answerA1
- B2
- C0
- DNone of these
Solution
$x \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$
and $x \sin \theta=y \cos \theta$
Equation (i) may be written as $x \sin \theta \cdot \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$
$\Rightarrow y \cos \theta \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta$
$\Rightarrow y \cos \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta \cos \theta$
$\Rightarrow y \cos \theta=\sin \theta \cos \theta \therefore y=\sin \theta \quad \ldots$
Putting the value of y from
(iii) in (ii), we get $\mathrm{x} \sin \theta=\sin \theta \cdot \cos \theta \Rightarrow \mathrm{x}=\cos \theta \quad \ldots$ (iv)
Squaring (iii) and (iv) and adding, we get $\mathrm{x}^{2}+\mathrm{y}^{2}=\cos ^{2} \theta+\sin ^{2} \theta=1$
Hence, the correct answer is option: 'A'.