Question

Please answer this question​

Answer 1

Differentiate x and y with respect to t :

`x = e^(\sec(2t)) \implies (dx)/(dt) = e^(\sec(2t)) (d(\sec(2t)))/(dx) = 2\sec(2t)\tan(2t) e^(\sec(2t))`

`y = e^(\tan(2t)) \implies (dy)/(dt) = e^(\tan(2t)) (d(\tan(2t)))/(dx) = 2\sec^2(2t) e^(\tan(2t))`

By the chain rule,

`(dy)/(dx) = (dy)/(dt) * (dt)/(dx) = ((dy)/(dt))/((dx)/(dt))`

Then

`(dy)/(dx) = (2\sec^2(2t) e^(\tan(2t)))/(2\sec(2t)\tan(2t) e^(\sec(2t))) = (\sec(2t) e^(\tan(2t)))/(\tan(2t) e^(\sec(2t))) =(y \sec(2t))/(x\tan(2t))`

and we have

`\log(x) = \sec(2t)`

`\log(y) = \tan(2t)`

and the claim follows.