Question

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

• A. $n^2$
• B. $2n^2$
• C. $3n^2$
• D. $4n^2$

Answer 1

The correct option is D $4n^2$

Given, $x=\sec \theta -\cos \theta$ and $y={\sec }^n\theta -{\cos }^n\theta$

We are supposed to find out of the value of $(x^2+4){\left(\frac{dy}{dx}\right)}^2-n^2{y}^2$

Now,

$x^2+4={(\sec \theta -\cos \theta )}^2+4$

$={\sec }^2\theta +{\cos }^2\theta -2\sec \theta .\cos \theta +4$

$={\sec }^2\theta +{\cos }^2\theta +2$ ($\because$ $\sec \theta .\cos \theta =1$)

$={\sec }^2\theta +{\cos }^2\theta +2\sec \theta .\cos \theta$

$={(\sec \theta +\cos \theta )}^2$

Also,

$y^2+4={({\sec }^n\theta -{\cos }^n\theta )}^2+4={({\sec }^n\theta +{\cos }^n\theta )}^2$

Now,

$\frac{dy}{d\theta }=n{\sec }^{n-1}\theta (\sec \theta .\tan \theta )-{n\cos }^{(n-1)}\theta .(-\sin \theta )$

$=n{\sec }^n\theta .\tan \theta +n{\cos }^{n-1}\theta .\sin \theta$

$=n\tan \theta ({\sec }^n\theta +{\cos }^n\theta )$

Similarly,

$\frac{dx}{d\theta }=\sec \theta .\tan \theta +\sin \theta =\tan \theta (\sec \theta +\cos \theta )$

$\therefore$ $\frac{dy}{dx}=\frac{n\tan \theta ({\sec }^n\theta +{\cos }^n\theta )}{\tan \theta (\sec \theta +\cos \theta )}$

$=\frac{n({\sec }^n\theta +{\cos }^n\theta )}{\sec \theta +\cos \theta }$

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

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

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

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

Answer : Option D.