Question
Write the least value of $\cos^2\text{x}+\sec^2\text{x}.$

Answer

$-1\le\cos\text{x}\le1$ $\Rightarrow1\le\cos^2\text{x}\le1$ $-1\le\frac{1}{\cos\text{x}}\le1$ $\Rightarrow1\le\frac{1}{\cos^2\text{x}}\le1$ $\Rightarrow1\le\sec^2\text{x}\le1$ $2\le\cos^2\text{x}+\sec^2\text{x}$. So the least value of $\cos^2\text{x}+\sec^2\text{x}$ is 2.

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