Limit Continuity And Differentiability Ques 24
- If $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$, then show that
$\left(x^2+4\right)\left(\frac{d y}{d x}\right)^2=n^2\left(y^2+4\right)$.
(1989, 2M)
Show Answer
Solution: Given, $x=\sec \theta-\cos \theta$ and $y=\sec ^n \theta-\cos ^n \theta$
On differentiating w.r.t. $\theta$ respectively, we get
$ \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta $
and $\frac{d y}{d \theta}=n \sec ^{n-1} \theta \cdot \sec \theta \tan \theta-n \cos ^{n-1} \theta \cdot(-\sin \theta)$
$\Rightarrow \quad \frac{d x}{d \theta}=\tan \theta(\sec \theta+\cos \theta)$
and $\frac{d y}{d \theta}=n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{n\left(\sec ^n \theta+\cos ^n \theta\right)}{\sec \theta+\cos \theta}$
$\therefore \quad \left(\frac{d y}{d x}\right)^2=\frac{n^2\left(\sec ^n \theta+\cos ^n \theta\right)^2}{(\sec \theta+\cos \theta)^2}$
$= \quad \frac{n^2\left\{\left(\sec ^n \theta-\cos ^n \theta\right)^2+4\right\}}{\left\{(\sec \theta-\cos \theta)^2+4\right\}}=\frac{n^2\left(y^2+4\right)}{\left(x^2+4\right)}$
$\Rightarrow \quad \left(x^2+4\right)\left(\frac{d y}{d x}\right)^2=n^2\left(y^2+4\right)$
BETA
