If $x = a\left(\cos t + \log \tan\frac{t}{2}\right), y = a\sin t$, then value of $\frac{dy}{dx}$ at $t = \frac{\pi}{4}$ is

Since both $x$ and $y$ are given in terms of a parameter $t$:

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

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$y = a\sin t$

$\dfrac{dy}{dt} = a\cos t$

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$x = a\left(\cos t + \log\tan\dfrac{t}{2}\right)$

$\dfrac{dx}{dt} = a\left(-\sin t + \dfrac{1}{\tan\frac{t}{2}} \cdot \sec^2\dfrac{t}{2} \cdot \dfrac{1}{2}\right)$

The second term inside the bracket simplifies as:

$\dfrac{\cos\frac{t}{2}}{\sin\frac{t}{2}} \cdot \dfrac{1}{\cos^2\frac{t}{2}} \cdot \dfrac{1}{2}$

$= \dfrac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}}$

$= \dfrac{1}{\sin t}$

So:

$\dfrac{dx}{dt} = a\left(-\sin t + \dfrac{1}{\sin t}\right)$

$= a\left(\dfrac{-\sin^2 t + 1}{\sin t}\right)$

$= a\left(\dfrac{\cos^2 t}{\sin t}\right)$

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$\dfrac{dy}{dx} = \dfrac{a\cos t}{\;a\cdot\dfrac{\cos^2 t}{\sin t}\;}$

$= \dfrac{\cos t \cdot \sin t}{\cos^2 t}$

$= \dfrac{\sin t}{\cos t}$

$= \tan t$

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At $t = \dfrac{\pi}{4}$:

$\dfrac{dy}{dx}\Bigg|_{t=\pi/4} = \tan\dfrac{\pi}{4} = \boxed{1}$