If $x=a\, sin\, t $ and $y = a(cost + log\, tan\frac{t}{2})$ then $\frac{d^2y}{dx}$ is :

$\frac{-1}{a}cosec^2tsec \, t $

The correct answer is Option (3) → $\frac{-1}{a}cosec^2t\sec t $

$x=a\sin t$

$y = a(\cos t + \log\tan\frac{t}{2})$

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

$\frac{dy}{dt}=-a\sin t+\frac{a\sec^2\frac{t}{2}}{2\tan\frac{t}{2}}$

$\frac{dy}{dt}=-a\sin t+\frac{a\cos \frac{t}{2}}{2\sin\frac{t}{2}\cos^2\frac{t}{2}}$

$=-a\sin t+\frac{a}{\sin t}=\frac{a(1-\sin^2t)}{\sin t}$

$\frac{dy}{dt}=\frac{a\cos^2t}{\sin t}$

so $\frac{dy}{dx}=\cot t⇒\frac{d^2y}{dx^2}=-cosec^2t\frac{dt}{dx}$

$=\frac{-cosec^2t\sec t}{a}$