NCERT Class 12 Mathematics Solutions – Exercise 2.3 Question 1
NCERT Class 12 Maths | Chapter 2 Inverse Trigonometric Functions
Question
Find the value of \[ \tan^{-1}\left(\tan \frac{5\pi}{6}\right) + \cos^{-1}\left(\cos \frac{13\pi}{6}\right) \]
[NCERT Ex 2.3, Question 1, Page 35]
Solution
We know that:
\[ \tan^{-1}(\tan x) = x,\; x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] \[ \cos^{-1}(\cos x) = x,\; x \in [0,\pi] \]
Therefore,
\[ \tan^{-1}\left(\tan \frac{5\pi}{6}\right) + \cos^{-1}\left(\cos \frac{13\pi}{6}\right) \]
\[ = \tan^{-1}\left(\tan\left(\pi – \frac{\pi}{6}\right)\right) + \cos^{-1}\left(\cos\left(2\pi + \frac{\pi}{6}\right)\right) \]
\[ = \tan^{-1}\left(-\tan\frac{\pi}{6}\right) + \cos^{-1}\left(\cos\frac{\pi}{6}\right) \]
Using identity: \[ \tan^{-1}(-x) = -\tan^{-1}x \]
\[ = -\frac{\pi}{6} + \frac{\pi}{6} \]
\[ \boxed{0} \]
