Question

If $2ta{n}^{-1}x=si{n}^{-1}\frac{2x}{1+x^2}$, then:

• A. $x\in R$
• B. $x\ge 1$
• C. $x\le -1$
• D. $-1\le x\le 1$

Answer 1

The correct option is D $-1\le x\le 1$

Putting $\theta =ta{n}^{-1}x,-\frac{\pi }{2}<\theta <\frac{\pi }{2}$, we have
R.H.S.
=$si{n}^{-1}\frac{2tan\theta }{1+ta{n}^2\theta }=si{n}^{-1}sin2\theta =2\theta =2ta{n}^{-1}x$
When $-\frac{\pi }{2}\le 2\theta \le \frac{\pi }{2}$
i.e., $-\frac{\pi }{4}\le \theta \le \frac{\pi }{4}$
i.e., $-1\le tan\theta \le 1$
i.e.,$-1\le x\le 1$