NCERT Class 12 Mathematics Solutions – Exercise 2.3 Question 13
NCERT Class 12 Maths | Chapter 2 Inverse Trigonometric Functions
Question
Find the simplified form of \[ \cos^{-1}\left(\frac{3}{5}\cos x + \frac{4}{5}\sin x\right), \quad x \in \left[-\frac{3\pi}{4}, \frac{\pi}{4}\right] \]
[NCERT EXEMPLAR Ex 2.3, Question 13, Page 36]
Solution
We have,
\[ \cos^{-1}\left(\frac{3}{5}\cos x + \frac{4}{5}\sin x\right) \]
Let \[ \cos y = \frac{3}{5} \quad \Rightarrow \quad \sin y = \frac{4}{5} \]
Therefore, \[ y = \cos^{-1}\frac{3}{5} = \sin^{-1}\frac{4}{5} = \tan^{-1}\left(\frac{4}{3}\right) \]
Now, \[ \cos^{-1}(\cos y \cos x + \sin y \sin x) \]
\[ = \cos^{-1}(\cos(y – x)) \]
\[ = y – x \]
\[ \boxed{\tan^{-1}\left(\frac{4}{3}\right) – x} \]
