The length of the tangent to the curve x = a( | vedclass

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(C) Given the curve $x = a(\cos t + \log 	an(t/2))$ and $y = a \sin t$.First,find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = a(-\sin t + \frac{1

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$a$

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(C) Given the curve $x = a(\cos t + \log 	an(t/2))$ and $y = a \sin t$.First,find the derivative $\frac{dy}{dx}$:$\frac{dx}{dt} = a(-\sin t + \frac{1}{	an(t/2)} \cdot \sec^2(t/2) \cdot \frac{1}{2}) = a(-\sin t + \frac{1}{2 \sin(t/2) \cos(t/2)}) = a(-\sin t + \frac{1}{\sin t}) = a(\frac{1-\sin^2 t}{\sin t}) = a \frac{\cos^2 t}{\sin t}$.$\frac{dy}{dt} = a \cos t$.Therefore,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \cos t}{a \cos^2 t / \sin t} = 	an t$.The slope of the tangent $m_T = 	an t$. Let the angle of inclination be $	heta$,so $	an 	heta = 	an t$,which implies $	heta = t$.The length of the tangent $PC$ is given by the formula $PC = |y \csc 	heta|$.Substituting $y = a \sin t$ and $	heta = t$:$PC = |a \sin t \cdot \csc t| = |a \sin t \cdot \frac{1}{\sin t}| = |a|$.Since $a$ is a constant length,the length of the tangent is $a$.

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

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