\(\text{ If }y(\theta )=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}\text{, then at …

Given: \(y(\theta )=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}\)

We know that, \(\cos 2\theta =2{\cos }^{2}\theta -1\)

\(\cos 3\theta =4{\cos }^{3}\theta -3\cos \theta\)

\(y(\theta )=\frac{2\cos \theta +2{\cos }^{2}\theta -1}{4{\cos }^{3}\theta -3\cos \theta +8{\cos }^{2}\theta -4+5\cos \theta +2}\)

\(y(\theta )=\frac{2{\cos }^{2}\theta +2\cos \theta -1}{4{\cos }^{3}\theta +8{\cos }^{2}\theta +2\cos \theta -2}\)

\(y(\theta )=\frac{2{\cos }^{2}\theta +2\cos \theta -1}{(2\cos \theta +2)(2{\cos }^{2}\theta +2\cos \theta -1)}\)

\(y(\theta )=\frac{1}{2(\cos \theta +1)}\)

\(y(\theta )=\frac{1}{2\cdot 2{\cos }^{2}\frac{\theta }{2}}\)

\(y(\theta )=\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\)

\(y'(\theta )=\frac{1}{4}\cdot 2\sec \frac{\theta }{2}\left(\sec \frac{\theta }{2}\tan \frac{\theta }{2}\right)\frac{1}{2}\)

\(y'(\theta )=\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\tan \frac{\theta }{2}\)

\(y''(\theta )=\frac{1}{4}\left({\sec }^{4}\frac{\theta }{2}\times \frac{1}{2}+{\sec }^{2}\frac{\theta }{2}{\tan }^{2}\frac{\theta }{2}\right)\)

\(y''(\theta )=\frac{1}{8}{\sec }^{4}\frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\frac{\theta }{2}{\tan }^{2}\frac{\theta }{2}\)

\(y''(\theta )+y'(\theta )+y(\theta )=\frac{1}{8}{\sec }^{4}\frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\\ \frac{\theta }{2}{\tan }^{2}\frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\tan \frac{\theta }{2}+\frac{1}{4}{\sec }^{2}\frac{\theta }{2}\)

\(\text{at }\theta =\frac{\pi }{2},\)

\(y''\left(\frac{\pi }{2}\right)+y'\left(\frac{\pi }{2}\right)+y'\left(\frac{\pi }{2}\right)=\frac{1}{8}\times 4+\frac{1}{4}\times 2\times 1+\frac{1}{4}\times 2\times 1+\frac{1}{4}\times 2\)

\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)