Question

please answer this question​

Answer 1

Differentiate x and y with respect to t :

$x = e^{\sec(2t)} \implies \dfrac{dx}{dt} = e^{\sec(2t)} \dfrac{d(\sec(2t))}{dx} = 2\sec(2t)\tan(2t) e^{\sec(2t)}$

$y = e^{\tan(2t)} \implies \dfrac{dy}{dt} = e^{\tan(2t)} \dfrac{d(\tan(2t))}{dx} = 2\sec^2(2t) e^{\tan(2t)}$

By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Then

$\dfrac{dy}{dx} = \dfrac{2\sec^2(2t) e^{\tan(2t)}}{2\sec(2t)\tan(2t) e^{\sec(2t)}} = \dfrac{\sec(2t) e^{\tan(2t)}}{\tan(2t) e^{\sec(2t)}} =\dfrac{y \sec(2t)}{x\tan(2t)}$

and we have

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

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

and the claim follows.