Q. If $\theta = \sin^{- 1}{x + \cos^{- 1}{x - \tan^{- 1}{x \geq 0,}}}$ then the smallest interval in which $\theta$ lies, is given by
A) $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{4}$
B) $- \frac{\pi}{4} \leq \theta \leq 0$
C) $0 \leq \theta \leq \frac{\pi}{4}$
D) $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$
Explanation
(d)
We have, $\theta = \sin^{- 1}{x + \cos^{- 1}{x - \tan^{- 1}x}}$
$= \frac{\pi}{2} - \tan^{- 1}{x = \cot^{- 1}x}$
Since, 0$\ \leq x \leq 1,\$therefore $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.