# If x = a sin t and y = a (cos t + log tan t/2 ), then find d^2y/dx^2. ​

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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

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

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

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