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If x = a sin t and y = a (cos t + log tan \(\frac{t}{2}\)), then find \(\frac{d^2y}{dx^2}.\)

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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

\(\frac{d^2y}{dx^2}=-cosec^2\,t.\frac{dt}{dx}\)

⇒ \(\frac{d^2y}{dx^2}=-cosec^2\,t.\frac{1}{a\,cos\,t}\) 

\(=\frac{-\,cosec^2\,t}{a\,cos\,t}\)

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