If x=a (cos t + log tant/2), y= a sin t, then value of dy/dx at t=(π)…

The correct answer is Option (2) → 1

Given: $x=a\big(\cos t+\log\tan\frac{t}{2}\big),\; y=a\sin t$

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

$\frac{d}{dt}\log\tan\frac{t}{2}=\frac{1}{\tan\frac{t}{2}}\cdot\sec^{2}\frac{t}{2}\cdot\frac{1}{2} =\frac{1}{2}\cdot\frac{1}{\sin\frac{t}{2}\cos\frac{t}{2}} =\frac{1}{\sin t}$

$\frac{dx}{dt}=a\big(-\sin t+\frac{1}{\sin t}\big)$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{a\cos t}{a\big(-\sin t+\frac{1}{\sin t}\big)} =\frac{\cos t}{-\sin t+\frac{1}{\sin t}}$

Hence

$\frac{dy}{dx}=\frac{\cos t}{\frac{\cos^{2}t}{\sin t}}=\frac{\sin t}{\cos t}=\tan t$

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

$\frac{dy}{dx}=\tan\frac{\pi}{4}=1$

Therefore $\displaystyle \frac{dy}{dx}\bigg|_{t=\pi/4}=1$.