Question

lf $x={\cos }^n\theta ,y={\sin }^n\theta$ then $\frac{d^{2}y}{d{x}^2}=$

• A. $\frac{n}{n-1}.\frac{{\cos }^{2n-1}\theta }{{\sin }^{n-3}\theta }$
• B. $\frac{n-2}{n}.\frac{{\tan }^{n-3}\theta }{{\cos }^{n+1}\theta \sin \theta }$
• C. $n-1.{\tan }^{n-2}\theta .{\sec }^2\theta$
• D. $n\frac{{\sin }^{n-1}\theta }{{\cos }^{n-2}\theta }$

Answer 1

The correct option is B $\frac{n-2}{n}.\frac{{\tan }^{n-3}\theta }{{\cos }^{n+1}\theta \sin \theta }$

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{n{\sin }^{n-1}\theta \cos \theta }{-n{\cos }^{n-1}\theta \sin \theta }$

$=-{\tan }^{n-2}\theta$

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

$=\frac{-\left(\right(n-2\left){\tan }^{n-3}\sin \theta \right)}{-n{\cos }^{n-1}\theta \sin \theta }$

$=\frac{(n-2)}{n}\frac{{\tan }^{n-3}\theta }{{\cos }^{n+1}\theta \sin \theta }$