Question

If $x=sec\theta -\cos \theta \mathrm{and}y=se{c}^n\theta -{\cos }^n\theta$, then

• A. $(x^2+4){\left(\frac{dy}{dx}\right)}^2=n^2(y^2+4)$
• B. $(x^2+4){\left(\frac{dy}{dx}\right)}^2=x^2(y^2+4)$
• C. $(x^2+4){\left(\frac{dy}{dx}\right)}^2=(y^2+4)$
• D. None of these

Answer 1

The correct option is A

$(x^2+4){\left(\frac{dy}{dx}\right)}^2=n^2(y^2+4)$

Explanation for the correct option.

Step 1: Differentiate $x\mathrm{and}y$ with respect to $\theta$.

$\begin{array}{rcl}x& =& sec\theta -\cos \theta \\ & \Rightarrow & \frac{dx}{d\theta }=sec\theta \tan \theta +\sin \theta \end{array}$

$\begin{array}{rcl}y& =& se{c}^n\theta -{\cos }^n\theta \\ & \Rightarrow & \frac{dy}{d\theta }=n\left(se{c}^{n-1}\theta \right)\left(sec\theta \tan \theta \right)+n\left({\cos }^{n-1}\theta \right)\sin \theta \\ & =& n\left(se{c}^n\theta \tan \theta \right)+n\left({\cos }^{n-1}\theta \right)\sin \theta \end{array}$

Step 2: Find $\frac{dy}{dx}$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{d\theta }\times \frac{d\theta }{dx}\\ & =& \frac{n\left(se{c}^n\theta \tan \theta \right)+n\left({\cos }^{n-1}\theta \right)\sin \theta }{sec\theta \tan \theta +\sin \theta }\\ & =& \frac{n\left(se{c}^n\theta \tan \theta +{\displaystyle \frac{{\cos }^n\theta }{\cos \theta }}\sin \theta \right)}{sec\theta \tan \theta +\tan \theta \cos \theta }\\ & =& \frac{n\tan \theta \left(se{c}^n\theta +{\cos }^n\theta \right)}{\tan \theta \left(sec\theta +\cos \theta \right)}\\ & =& \frac{n\left(se{c}^n\theta +{\cos }^n\theta \right)}{\left(sec\theta +\cos \theta \right)}\end{array}$

Step 3: Find ${\left(\frac{dy}{dx}\right)}^2$

$$\begin{gathered} {\left(\frac{dy}{dx}\right)}^2={\left[\begin{array}{rcl}& & \frac{n\left(se{c}^n\theta +{\cos }^n\theta \right)}{\left(sec\theta +\cos \theta \right)}\end{array}\right]}^2 \\ =\frac{n^2\left[{\begin{array}{rcl}& & \left(se{c}^n\theta -{\cos }^n\theta \right)\end{array}}^2+4\right]}{\begin{array}{rcl}& & \left(sec\theta -\cos \theta \right)\end{array}\begin{array}{rcl}& & \end{array}} \end{gathered}$$

Hence, option A is correct.