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If $y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}$, then at $\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y$ is equal to :
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$\begin{aligned} & y=\frac{2 \cos \theta+2 \cos ^2 \theta-1}{4 \cos ^3 \theta-3 \cos \theta+8 \cos ^2 \theta-4+5 \cos \theta+2} \\ & y=\frac{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & y=\frac{1}{2}\left(\frac{1}{1+\cos \theta}\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2} \\ & y^{\prime}=\frac{1}{2}\left(\frac{-1}{(1+\cos \theta)^2} \times(-\sin \theta)\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2}\end{aligned}$
$\begin{aligned} y^{\prime \prime} & =\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^2-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^4}\right] \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=1\end{aligned}$
$\begin{aligned} y^{\prime \prime} & =\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^2-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^4}\right] \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=1\end{aligned}$
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