Question

Find the minimum value of ${\cos }^2\theta +$ ${\sec }^2\theta$

Answer 1

Let $f\left(\theta \right)=y={\cos }^2\theta +{\sec }^2\theta =\frac{{\cos }^4\theta +1}{{\cos }^2\theta }$

For $\frac{max}{{min}_1}$

$\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-{\cos }^2\theta \left(4{\cos }^3\theta \right)\sin \theta +2\cos \theta \sin \theta ({\cos }^4\theta +1)}{{({\cos }^4\theta +1)}^2}=0$

So, $2\cos \theta \sin \theta [1+{\cos }^4\theta -2{\cos }^4\theta ]=0$

$2{\cos }^4\theta \sin \theta (1-{\cos }^4\theta )=0$

$\sin \theta =0,\cos \theta =0$

So, $\cos \theta =\pm 1$

So, $\theta =n\pi ,\theta =(2n+1)\frac{\pi }{2},\theta =n\pi$

$f\left(n\pi \right)=\frac{1+1}{1}=2$

$f\left((2n+1)\frac{\pi }{2}\right)=\frac{0+1}{0}=\infty$

So, minima arrives at $\theta =n\pi$

Thus, minimum value$=2$