Question

If $\theta \equiv si{n}^{-1}x+co{s}^{-1}x-ta{n}^{-1}x,0\le x\le 1$, then the smallest interval in which $\theta$ lies is given by ?

• A. $\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$
• B. $-\frac{\pi }{4}\le \theta \le 0$
• C. $0\le \theta \le \frac{\pi }{4}$
• D. $\frac{\pi }{2}\le \theta \le \frac{3\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$

Given $0\le x\le 1⟹{\tan }^{-1}x\in [{\tan }^{-1}0,{\tan }^{-1}1]⟹{\tan }^{-1}x\in [0,\frac{\pi }{4}]⟹-{\tan }^{-1}x\in [-\frac{\pi }{4},0]$

$\theta ={\sin }^{-1}x+{\cos }^{-1}x-{\tan }^{-1}x=\frac{\pi }{2}-{\tan }^{-1}x$ $(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2})$

As $-{\tan }^{-1}x\in [-\frac{\pi }{4},0]$

$⟹\frac{\pi }{2}-{\tan }^{-1}x\in [\frac{\pi }{4},\frac{\pi }{2}]$

$⟹\frac{\pi }{4}\le \theta \le \frac{\pi }{2}$