NCERT Class 12 Mathematics Solutions – Exercise 2.3 Question 4
NCERT Class 12 Maths | Chapter 2 Inverse Trigonometric Functions
Question
Find the value of \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right] \]
[NCERT Ex 2.3, Question 4, Page 35]
Solution
We have:
\[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) + \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) + \tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right] \]
\[ = \tan^{-1}\left(\tan\frac{5\pi}{6}\right) + \cot^{-1}\left(\cot\frac{\pi}{3}\right) + \tan^{-1}(-1) \]
\[ = \tan^{-1}\left(\tan\left(\pi – \frac{\pi}{6}\right)\right) + \cot^{-1}\left(\cot\frac{\pi}{3}\right) + \tan^{-1}\left(\tan\left(\pi – \frac{\pi}{4}\right)\right) \]
\[ = \tan^{-1}\left(-\tan\frac{\pi}{6}\right) + \cot^{-1}\left(\cot\frac{\pi}{3}\right) + \tan^{-1}\left(-\tan\frac{\pi}{4}\right) \]
\[ = -\frac{\pi}{6} + \frac{\pi}{3} – \frac{\pi}{4} \]
\[ = \frac{-2\pi + 4\pi – 3\pi}{12} \]
\[ \boxed{-\frac{\pi}{12}} \]