Question

If $x = \sec \theta - \cos \; \theta, y = \sec^n \theta - \cos^n \theta $, then $(x^2 + 4) \left( \frac{dy}{dx} \right)^2$ is equal to

• A. $n^2 (y^2 -4)$
• B. $n^2 (4 - y^2 )$
• C. $n^2 (y^2 + 4)$
• D. $None\, of \,these$

Answer 1

Answer: (C)

$x=\sec \theta-\cos \theta$
$\Rightarrow \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta,$
$y=\sec ^{n} \theta-\cos ^{n} \theta$
$\Rightarrow \frac{d y}{d \theta}=n \sec ^{n-1} \theta \sec \theta \tan \theta +n \cos ^{n-1} \theta \sin \theta$
$\therefore \frac{d y}{d x}=n \frac{\left(\sec ^{n} \theta \tan \theta+\cos ^{n-1} \theta \sin \theta\right)}{(\sec \theta \tan \theta+\sin \theta)}$
$\Rightarrow \frac{d y}{d x}=n \frac{\left(\sec ^{n} \theta+\cos ^{n} \theta\right) \tan \theta}{(\sec \theta+\cos \theta) \tan \theta}$
$\Rightarrow \frac{d y}{d x}=\frac{n\left(\sec ^{n} \theta+\cos ^{n} \theta\right)}{(\sec \theta+\cos \theta)}$
$\Rightarrow \left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left\{\left(\sec ^{n} \theta-\cos ^{n} \theta\right)^{2}+4\right\}}{(\sec \theta-\cos \theta)^{2}+4}$
$\Rightarrow \left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left(y^{2}+4\right)}{x^{2}+4}$
$\Rightarrow \left(x^{2}+4\right)\left(\frac{d y}{d x}\right)^{2}=n^{2}\left(y^{2}+4\right)$