`x=tcost ,y=t+sintdot` Then `(d^(2x))/(dy^2)at t=pi/2` is `(a)(pi+4)/2` (b) `-(pi+4)/2` `(c) -2` (d) none of these

Options

A. `(pi+4)/(2)`
B. `-(pi+4)/(2)`
C. `-2`
D. none of these

Explanation

`(dx)/(dy)=((dx)/(dt))/((dy)/(dt))=(cos t - t sin t)/(1+ cos t)`
`therefore" "(d^(2)x)/(dy^(2))=((d)/(dt)((dx)/(dy)))/((dy)/(dt))`
`=(((-2 sin t- t cos t)(1+ cos t )-(cos t - t sin t)(- sin t))/((1+cos t)^(2)))/(1+cos t)`