Question

Solve this:


$$\begin{gathered} 39.\mathrm{If}\mathrm{x}\ge 0\mathrm{and}\mathrm{\theta }={\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}-{\tan }^{-1}\mathrm{x},\mathrm{then} \\ \left(1\right)\frac{\mathrm{\pi }}{2}\le \mathrm{\theta }\le \frac{3\mathrm{\pi }}{4}\left(2\right)0<\mathrm{\theta }\le \frac{\mathrm{\pi }}{4} \\ \left(4\right)0<\mathrm{\theta }\le \frac{\mathrm{\pi }}{3}\left(4\right)\frac{\mathrm{\pi }}{4}\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2} \end{gathered}$$

Answer 1

$$\begin{gathered} x\ge 0and\theta ={\sin }^{-1}x+{\cos }^{-1}x-{\tan }^{-1}x \\ Weknow \\ {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2} \\ \theta =\frac{\mathrm{\pi }}{2}-{\tan }^{-1}x \\ Sincex\ge 0 \\ so \\ 0\le {\tan }^{-1}x<\frac{\mathrm{\pi }}{2} \\ 0\ge -{\tan }^{-1}x>-\frac{\mathrm{\pi }}{2} \\ \frac{\mathrm{\pi }}{2}\ge \frac{\mathrm{\pi }}{2}-{\tan }^{-1}x>\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2} \\ \frac{\mathrm{\pi }}{2}\ge \theta >0 \\ 0<\theta \le \frac{\mathrm{\pi }}{2} \end{gathered}$$