Inverse Trigonometric Functions — Class 12 Maths Solution

ncert exercise SA NCERT Ex. 2.2, Q. 19 , Page 48
Question

${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right)$ is equal to

(A) $\frac{{7\pi }}{6}$

(B) $\frac{{5\pi }}{6}$

(C) $\frac{\pi }{3}$

(D) $\frac{\pi }{6}$

Step-by-step Solution

Option B is correct

${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) \ne \frac{{7\pi }}{6}$ as principal value branch of ${\cos ^{ - 1}}\;\,is\;\;[0,\;\pi ]$

So, ${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left[ {\cos \left( {\pi + \frac{\pi }{6}} \right)} \right]$

$= {\cos ^{ - 1}}\left( { - \cos \frac{\pi }{6}} \right) = {\cos ^{ - 1}}\left( {\cos \left( {\pi - \frac{\pi }{6}} \right)} \right) = \frac{{5\pi }}{6}.$

Hence, ${\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) = \frac{{5\pi }}{6}.$

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