Inverse Trigonometric Functions — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 2.1, Q. 13 , Page 42
Question

. 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) $- \frac{\pi }{2} < y < \frac{\pi }{2}$

Step-by-step 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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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Inverse Trigonometric Functions. Curated by Sachin Sharma. Free for all students.