Given,
x = a sin t
Differentiating both sides with respect to t, we get
\(\frac{dx}{dt}=\) a cos t ...(i)
Again,
∵ y = a [ cos t + log (tan\(\frac{t}{2}\))]
Differentiating both sides with respect to t, we get
Differentiating both sides with respect to t, we get
\(\frac{d^2y}{dx^2}=-cosec^2\,t.\frac{dt}{dx}\)
⇒ \(\frac{d^2y}{dx^2}=-cosec^2\,t.\frac{1}{a\,cos\,t}\)
\(=\frac{-\,cosec^2\,t}{a\,cos\,t}\)