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