If x = sec theta - cos theta and y = sec^n th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. Let $u = \sec 	heta$ and $v = \cos 	heta = \frac{1}{u}$. Then $x

Vedclass · Accepted Answer

$\frac{{n^2(y^2 + 4)}}{{x^2 + 4}}$

Vedclass · Answer

(A) Given $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$. Let $u = \sec 	heta$ and $v = \cos 	heta = \frac{1}{u}$. Then $x = u - \frac{1}{u}$ and $y = u^n - \frac{1}{u^n}$. We know that $x^2 = (u - \frac{1}{u})^2 = u^2 + \frac{1}{u^2} - 2$,so $x^2 + 4 = u^2 + \frac{1}{u^2} + 2 = (u + \frac{1}{u})^2$. Thus,$u + \frac{1}{u} = \sqrt{x^2 + 4}$. Similarly,$y^2 + 4 = (u^n - \frac{1}{u^n})^2 + 4 = u^{2n} + \frac{1}{u^{2n}} - 2 + 4 = u^{2n} + \frac{1}{u^{2n}} + 2 = (u^n + \frac{1}{u^n})^2$. Thus,$u^n + \frac{1}{u^n} = \sqrt{y^2 + 4}$. Using the identity $(u^n - \frac{1}{u^n})^2 = (u^n + \frac{1}{u^n})^2 - 4$,we have $y^2 = (y^2 + 4) - 4$. Now,differentiate $x$ and $y$ with respect to $	heta$: $\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta = \sec 	heta 	an 	heta + \sin 	heta = 	an 	heta (\sec 	heta + \cos 	heta)$. $\frac{dy}{d	heta} = n \sec^{n-1} 	heta (\sec 	heta 	an 	heta) + n \cos^{n-1} 	heta \sin 	heta = n 	an 	heta (\sec^n 	heta + \cos^n 	heta)$. Then $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{n(\sec^n 	heta + \cos^n 	heta)}{\sec 	heta + \cos 	heta} = \frac{n \sqrt{y^2 + 4}}{\sqrt{x^2 + 4}}$. Squaring both sides,we get ${\left( {\frac{{dy}}{{dx}}} ight)^2} = \frac{{n^2(y^2 + 4)}}{{x^2 + 4}}$.

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then ${\left( {\frac{{dy}}{{dx}}} \right)^2}$ equals

Similar Questions

If $x = a(t - \frac{1}{t})$ and $y = a(t + \frac{1}{t})$,then $\frac{dy}{dx} = $

If $y = \sin \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)$ and $x = \cos 2 \theta$,then $\frac{d y}{d x} =$

If,for $a \neq 0$,$x = a(1 - \sin t)$ and $y = a(t + \cos t)$,then $\frac{d^2 y}{d x^2} = $

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$,then find the value of $\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$.

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module