Inverse Trigonometric Functions — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 2.1, Q. 3 , Page 41
Question

${\rm{cose}}{{\rm{c}}^{ - 1}}(2)$

Step-by-step Solution

Let ${\rm{cose}}{{\rm{c}}^{ - 1}}(2) = x \Rightarrow 2 = {\rm{cosec}}\;{\rm{x}}$

As we know that the range of the principal value branch of ${\rm{cose}}{{\rm{c}}^{ - 1}}\;\;is\;\;\left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right] - \{ 0\}$

Then, $2 = {\rm{cosec}}\;{\rm{x}}\;{\rm{ = cosec }}\left( {\frac{\pi }{6}} \right),\;where\;\frac{\pi }{6} \in \left[ {\frac{{ - \pi }}{2},\;\frac{\pi }{2}} \right] - \{ 0\}$

Hence, the principal value of ${\rm{cose}}{{\rm{c}}^{ - 1}}(2)\;is\;\pi /6.$

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