Question

If $x=\sin \theta ,y=\cos 3\theta$, then $(1-x^2)y_2-x{y}_1$ is equal to

• A. $y$
• B. $-3y$
• C. $3y$
• D. $-9y$

Answer 1

The correct option is D $-9y$

The given information is: $x=\sin \theta$, $y=\cos 3\theta$

Differentiating $x$ and $y$ w.r.t $\theta$ we get,

$\Rightarrow \frac{dx}{d\theta }=\cos \theta$

$\Rightarrow \frac{dy}{d\theta }=-3\sin \theta$

$\Rightarrow \frac{dy}{dx}=\frac{-3\sin \theta }{\cos \theta }$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left(\frac{dy}{dx}\right).\frac{d\theta }{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left(\frac{-3\sin \theta }{\cos \theta }\right).\frac{d\theta }{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-3(3\cos 3\theta \cos \theta -\sin \theta \sin 3\theta )}{{\cos }^2\theta }.\frac{1}{\cos \theta }$

$\Rightarrow (1-x^2)y_2=(1-{\sin }^2\theta )\frac{-3(3\cos 3\theta \cos \theta -\sin \theta \sin 3\theta )}{{\cos }^2\theta }.\frac{1}{\cos \theta }$

$\Rightarrow (1-x^2)y_2={\cos }^2\theta .\frac{-3(3\cos 3\theta \cos \theta -\sin \theta \sin 3\theta )}{{\cos }^2\theta }.\frac{1}{\cos \theta }$

$\Rightarrow (1-x^2)y_2=\frac{(-9\cos 3\theta \cos \theta +3\sin \theta \sin 3\theta )}{\cos \theta }$

$\Rightarrow x{y}_1=\frac{-3\sin \theta \sin 3\theta }{\cos \theta }$

$\Rightarrow (1-x^2)y_2+x{y}_1=\frac{-9\cos 3\theta \cos \theta +3\sin \theta \sin 3\theta -3\sin \theta \sin 3\theta }{\cos \theta }$

$\Rightarrow (1-x^2)y_2+x{y}_1=-9\cos 3\theta$

$\Rightarrow (1-x^2)y_2+x{y}_1=-9y$