Question

Find the range of $f\left(x\right)=\sqrt{4-\sqrt{1+{\tan }^{2}x}}$

Answer 1

$y=\sqrt{4-\sqrt{1+{\tan }^{2}x}}\Rightarrow y^2=4-\sqrt{{\sec }^{2}x}\Rightarrow {\sec }^{2}x=4-y^2\Rightarrow \sec x=\sqrt{4-y^2}\Rightarrow x={\sec }^{-1}\sqrt{4-y^2}$
As $\sec \theta \in (-\infty ,1)\cup (1,\infty ))$
Then $f\left(x\right)\in (4,\infty )$