If $x = \frac{1 - \cos^2t}{\cos t}$ and $y = \frac{1-\cos^{2n}t}{\cos^nt}$ find $\frac{dy}{dx}$.

Question

This is the given question and I have to prove $\left(\frac{dy}{dx}\right)^2 = \frac{n^2(y^2+4)}{x^2+4}$.

Please help me solve this, I need it ASAP.

I have tried, but I couldn't do it

Question

Answer 1

$$x^2=(\sec t- \cos t)^2=(\sec t+ \cos t)^2-4~~~~(1)$$ And $$y^2=(sec^n t- \cos^n t)^2=(\sec^{n}t+\cos^{n} t)^2-4~~~~(2)$$ $$\frac{dy}{dt}= n\tan t(\sec^{n} t+ \cos^{n} t) ~~~(3)$$ $$\frac{dx}{dt}=\tan t~(\sec t+\cos t)$$ So $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}= n \left(\frac{\sec^n t+ \cos^n t}{\sec t+ \cos t}\right)$$ $$\implies \left(\frac{dy}{dx}\right)^2=n^2 \left(\frac{\sec^n t+ \cos^n t}{\sec t+ \cos t}\right)^2= n^2 \frac{y^2+4}{x^2+4}$$

For OP: $$\frac{dy}{dt}=n \sec^{n-1} t ~\sec t \tan t+ n \cos^{n-1} t~ \sin t= n \tan t~ (\sec ^n t+ \cos^n t)$$

Answer 2

Hint:

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\dot y(t)}{\dot x(t)}.$$