Question

If $x=a{\sec }^3\theta \mathrm{and}y=a{\tan }^3\theta ,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{at}\theta =\frac{\pi }{3}\mathrm{is}$

• A. $\frac{1}{2}$
• B. 1
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

We have, $x=a{\sec }^3\theta \mathrm{and}y=a{\tan }^3\theta \mathrm{Differentiating}w.r.t.\theta ,\mathrm{we}\mathrm{get}\frac{\mathrm{dx}}{\mathrm{d\theta }}=3a{\sec }^2\theta \frac{d}{\mathrm{d\theta }}\left(\sec \theta \right)=3a{\sec }^3\theta \tan \theta \frac{\mathrm{dy}}{\mathrm{d\theta }}=3a{\tan }^2\theta \frac{d}{\mathrm{d\theta }}\left(\tan \theta \right)=3a{\sec }^2\theta {\tan }^2\theta \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d\theta }}}{\frac{\mathrm{dx}}{\mathrm{d\theta }}}=\frac{3a{\sec }^2\theta {\tan }^2\theta }{3a{\sec }^3\theta \tan \theta }=\frac{\tan \theta }{\sec \theta }=\sin \theta \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\theta =\frac{\pi }{3}}=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$