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(A) Given the parametric equations of the curve: $x = a \cos^3 t$ and $y = a \sin^3 t$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} =

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$x \sec t + y \csc t = a$

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(A) Given the parametric equations of the curve: $x = a \cos^3 t$ and $y = a \sin^3 t$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = 3a \cos^2 t (-\sin t) = -3a \cos^2 t \sin t$$\frac{dy}{dt} = 3a \sin^2 t (\cos t) = 3a \sin^2 t \cos t$Now,find the slope of the tangent $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\sin t}{\cos t} = -	an t$The equation of the tangent at point $(a \cos^3 t, a \sin^3 t)$ is:$y - a \sin^3 t = -\frac{\sin t}{\cos t} (x - a \cos^3 t)$Multiply both sides by $\cos t$:$y \cos t - a \sin^3 t \cos t = -x \sin t + a \cos^3 t \sin t$Rearrange the terms:$x \sin t + y \cos t = a \sin t \cos^3 t + a \sin^3 t \cos t$$x \sin t + y \cos t = a \sin t \cos t (\cos^2 t + \sin^2 t)$Since $\cos^2 t + \sin^2 t = 1$,we get:$x \sin t + y \cos t = a \sin t \cos t$Divide both sides by $\sin t \cos t$:$\frac{x \sin t}{\sin t \cos t} + \frac{y \cos t}{\sin t \cos t} = \frac{a \sin t \cos t}{\sin t \cos t}$$x \sec t + y \csc t = a$

What is the equation of the tangent to the curve $x = a \cos^3 t, y = a \sin^3 t$ at point $t$?

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