​If x=sinΘ and y=cosΘ, then what is the value of yy'' +  (y’)​2?

Dear Student,
Please find below the solution to the asked query:

y' means dydx and y"  means d2ydx2x=sinθDifferentiate with respect to θ,dx=cosθ...iy=cosθdy=-sinθ....iiii/idydx=-sinθcosθy'=dydx=-cotθ....iiiy"=d2ydx2=ddxdydx=ddxdydx =ddydx dx=d-cotθ 1cosθ  By i and iii=--cosec2θ1cosθy"=cosec2θ.1cosθNowyy"+y'2=cosθ.cosec2θ.1cosθ+-cotθ2yy"+y'2=cosec2θ+cot2θ

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