If y(θ)= 2 cos θ+cos 2 θ cos 3 θ+4 cos 2 θ+5 cos θ+2, then at θ= π 2, y^prime prime+y^prime+y is equal to :

$\begin{aligned} & y=\frac{2 \cos \theta+2 \cos ^2 \theta-1}{4 \cos ^3 \theta-3 \cos \theta+8 \cos ^2 \theta-4+5 \cos \theta+2} \\ & y=\frac{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & y=\frac{1}{2}\left(\frac{1}{1+\cos \theta}\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2} \\ & y^{\prime}=\frac{1}{2}\left(\frac{-1}{(1+\cos \theta)^2} \times(-\sin \theta)\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2}\end{aligned}$​y=4cos3θ−3cosθ+8cos2θ−4+5cosθ+22cosθ+2cos2θ−1​y=(2cos2θ+2cosθ−1)(2cosθ+2)(2cos2θ+2cosθ−1)​y=21​(1+cosθ1​)⇒θ=2π​y=21​y′=21​((1+cosθ)2−1​×(−sinθ))⇒θ=2π​y=21​​ $\begin{aligned} y^{\prime \prime} & =\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^2-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^4}\right] \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=1\end{aligned}$y′′​=21​[(1+cosθ)4cosθ(1+cosθ)2−sinθ(2)(1+cosθ)(−sinθ)​]⇒θ=2π​y=1​