Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

A: $2{\sin }^{-1}x-\pi /2=\pi /6$, so $x=\sin (\pi /3)=\sqrt{3}/2$, positive and below 1. B: ${\sec }^2\left({\tan }^{-1}2\right)=5$ and ${\csc }^2\left({\cot }^{-1}3\right)=10$; total 15. C: solve $2{\sin }^{-1}x-\pi /2={\sin }^{-1}(3x-2);x=1$ satisfies and is the largest possible $x$ in ${\sin }^{-1}x$ 's domain, hence $x\ge 1$ and >0. D: ${\sin }^{-1}(\sin 5)=5-2\pi$, which lies in $(-\pi /2,0)$, negative and less than 1.