What is the length of the tangent to the curv | vedclass

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(A) Given the parametric equations: $x = a(\cos t + \log 	an(t/2))$ and $y = a \sin t$.First,find the derivatives with respect to $t$:$\frac{dx}{dt}

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$a$

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(A) Given the parametric equations: $x = a(\cos t + \log 	an(t/2))$ and $y = a \sin t$.First,find the derivatives with respect to $t$:$\frac{dx}{dt} = a(-\sin t + \frac{1}{	an(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}) = a(-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}) = a(-\sin t + \frac{1}{\sin t}) = a(\frac{1 - \sin^2 t}{\sin t}) = a \frac{\cos^2 t}{\sin t}$.$\frac{dy}{dt} = a \cos t$.Now,find the slope of the tangent $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \cos^2 t / \sin t} = \frac{\sin t}{\cos t} = 	an t$.The formula for the length of the tangent is $L = |y \frac{\sqrt{1 + (dy/dx)^2}}{dy/dx}|$.Substituting the values: $L = |a \sin t \cdot \frac{\sqrt{1 + 	an^2 t}}{	an t}| = |a \sin t \cdot \frac{\sec t}{	an t}| = |a \sin t \cdot \frac{1}{\cos t} \cdot \frac{\cos t}{\sin t}| = |a| = a$ (assuming $a > 0$).

What is the length of the tangent to the curve $x = a(\cos t + \log \tan(t/2))$,$y = a \sin t$ at any point?

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