$\displaystyle\frac{dy}{dx}$ at $\displaystyle t=\frac{\pi}{4}$ for $\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$ and $y=a\sin{t}$ is

$\displaystyle x=\frac{\pi}{4}$ for $\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$ and

$y=a\sin{t}$

Differentiating w.r.t $t$, we get

$\displaystyle \frac { dx }{ dt } =a\left[ \frac { d }{ dt } \cos { t } +\frac { 1 }{ 2 } \frac { d }{ dt } \left( \log { \tan ^{ 2 }{ \frac { t }{ 2 } } } \right) \right]$

$\displaystyle\frac{dx}{dt}=a\left[-\sin{t}+\frac{1}{\displaystyle\tan{\frac{t}{2}}}\sec^2{\frac{t}{2}}\times\frac{1}{2}\right]$

$\displaystyle \frac { dx }{ dt } =a\left[ -\sin { t } +\frac { 1 }{ 2\displaystyle \frac { \sin { { t }/{ 2 } } }{ \cos { { t }/{ 2 } } } .\cos ^{ 2 }{ { t }/{ 2 } } } \right]$

$\displaystyle =a\left[-\sin{t}+\frac{1}{\displaystyle 2\sin{\frac{t}{2}}\cos{\frac{t}{2}}}\right]$

$\displaystyle =a\left[-\sin{t}+\frac{1}{\sin{t}}\right]$

$=a\left[ \displaystyle \frac { 1-\sin ^{ 2 }{ t } }{ \sin { t } } \right] =a\left[ \displaystyle \frac { \cos ^{ 2 }{ t } }{ \sin { t } } \right]$

$\displaystyle\frac{dy}{dt}=a\cos{t}$

$\displaystyle\therefore\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{a\cos{t}}{\displaystyle\frac{a\cos^2{t}}{\sin{t}}}=\tan{t}$

At $\displaystyle t=\frac{\pi}{4}$, $\displaystyle\frac{dy}{dx}=1$