Question

Solve ${\sec }^{-1}x>cose{c}^{-1}x$

Answer 1

Given, ${\sec }^{-1}x>{\csc }^{-1}x\Rightarrow \sin \left({\sec }^{-1}x\right)>\sin \left({\csc }^{-1}x\right)$
$\Rightarrow \sin \left({\sin }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)\right)>\sin \left({\sin }^{-1}\left(\frac{1}{x}\right)\right)$
$\Rightarrow \frac{1}{\sqrt{1-x^2}}>\frac{1}{x}\Rightarrow x>\sqrt{1-x^2}$ (for $(x>0)$)
$\Rightarrow x^2>1-x^2\Rightarrow 2x^2>1\Rightarrow x\in (-\infty ,-\frac{1}{\sqrt{2}})\cup (\frac{1}{\sqrt{2}},\infty )$
$\therefore x\in (\frac{1}{\sqrt{2}},\infty )$