MCQ
If $x$ takes non-positive permissible value, then ${\sin ^{ - 1}}x =$
- A${\cos ^{ - 1}}\sqrt {1 - {x^2}} $
- ✓$ - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
- C${\cos ^{ - 1}}\sqrt {{x^2} - 1} $
- D$\pi - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
Since $ - 1 \le x \le 0,$ therefore $\frac{{ - \pi }}{2} \le {\sin ^{ - 1}}x \le 0$
and so $\frac{{ - \pi }}{2} \le y \le 0$
We have $\cos y = \sqrt {1 - {{\sin }^2}y} $
$ \Rightarrow \,\,\cos y = \sqrt {1 - {x^2}} $, for $0 \le y \le \pi $ …..$(i)$
Now $ - \frac{\pi }{2} \le y \le 0\,\, \Rightarrow \,\,\frac{\pi }{2} \ge - y \ge 0$
$ \Rightarrow \,\,\cos \,\left( { - y} \right) = \sqrt {1 - {x^2}} $ {from $(i)$}
$ \Rightarrow \,\, - y = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \,\, $
$\Rightarrow \,\,y = - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.