. If ${\sin ^{ - 1}}x = y,$ then
(A) $0 \le y \le \pi$
(B) $- \frac{\pi }{2} \le y \le \frac{\pi }{2}$
(C) $0 < y < \pi$
(D) $- \frac{\pi }{2} < y < \frac{\pi }{2}$
. If ${\sin ^{ - 1}}x = y,$ then
(A) $0 \le y \le \pi$
(B) $- \frac{\pi }{2} \le y \le \frac{\pi }{2}$
(C) $0 < y < \pi$
(D) $- \frac{\pi }{2} < y < \frac{\pi }{2}$
Official Solution
Option B is correct
${\sin ^{ - 1}}x = y$
$\Rightarrow$ $x = \sin y,$ where the range of principal value branch of
${\sin ^{ - 1}}\;is\;\left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right],\;then\;y \in \left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right] \Rightarrow - \frac{\pi }{2} \le y \le \frac{\pi }{2}.$
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