Question

If $\theta =si{n}^{-1}x+co{s}^{-1}x+ta{n}^{-1}x,0\le x\le 1$, 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 D $\frac{\pi }{2}\le \theta \le \frac{3\pi }{4}$

$\theta =si{n}^{-1}x+co{s}^{-1}x+ta{n}^{-1}x$
Now, $si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}$
$\Rightarrow \theta =\frac{\pi }{2}+ta{n}^{-1}x$
For ,$0\le x\le 1$
$0\le ta{n}^{-1}x\le \frac{\pi }{4}$
$\Rightarrow \frac{\pi }{2}\le \theta \le \frac{3\pi }{4}$
Hence, option D is correct.