a^2 + 2 + \frac{1}{a^2} + b^2 + 2 + \frac{1}{b^2}
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- To find the minimum value of the expression \((\cos x + \sec x)^2 + (\sin x + \csc x)^2\), let us start by rewriting \(\sec x\) and \(\csc x\) in terms of \(\cos x\) and \(\sin x\):
- Let \(a = \cos x\) and \(b = \sin x\). Then:
- This expands to:
- Since \(a^2 + b^2 = \cos^2 x + \sin^2 x = 1\), the expression simplifies to:
- + 4 + \frac{1}{a^2} + \frac{1}{b^2} = 5 + \frac{1}{a^2} + \frac{1}{b^2}
- We apply the AM-GM inequality to \(\frac{1}{a^2} + \frac{1}{b^2}\):
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