Question

If $x=\cos ec\theta -\sin \theta ,y=\cos e{c}^n\theta -{\sin }^n\theta$ 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 $4n^2$

$x=cosec\theta -sin\theta$

$x^2+4=(cosec\theta -sin\theta {)}^2+4$

$=(cosec\theta +sin\theta {)}^2$

$y=cose{c}^n\theta -si{n}^n\theta$

$y^2+4=(cose{c}^n\theta +si{n}^n\theta {)}^2$

$\frac{dy}{d\theta }=n.cose{c}^n\theta (-cosec\theta cot\theta )+nsi{n}^{n-1}\theta .cos\theta .$

$=ncot\theta (cose{c}^n\theta +si{n}^n\theta )$

$\frac{dx}{d\theta }=cot\theta cosec\theta -cos\theta$

$=-cot\theta =(cosec\theta +sin\theta )$

$\frac{dy}{dx}=\frac{ncot\theta }{-cot\theta }\left(\frac{cose{c}^n\theta +si{n}^n\theta }{cosec\theta +sin\theta }\right)$

$(\frac{dy}{dx}{)}^2=\frac{n^2(cose{c}^n\theta +si{n}^n\theta {)}^2}{(-1{)}^2(cosec\theta +sin\theta {)}^2}$

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

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

option D