NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions
Thank you for reading this post, don't forget to subscribe!Exercise 2.1
Find the principal values of the following:
\(\quad 1. \sin^{-1}\left(\frac{-1}{2}\right)\)
\(\quad 2.\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
\(\quad 3. \csc^{-1}(2)\)
\(\quad 4. \tan^{-1}(-\sqrt{3})\)
\(\quad 5. \cos^{-1}(-\frac{1}{2})\)
\(\quad 6. \tan^{-1}(-1)\)
\(\quad 7. \tan^{-1}(\frac{2}{\sqrt{3}})\)
\(\quad 8. \cot^{-1}(\sqrt{3})\)
\(\quad 9. \cos^{-1}(-\frac{1}{\sqrt{2}})\)
\(\quad 10. \csc^{-1}(-\sqrt{2})\)
Find the values of the following:
11. \(\tan^{-1}(1) + \cos^{-1}(\frac{-1}{2}) + \sin^{-1}(-\frac{1}{2})\)
12. \(\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2}\)
13. If \(\sin^{-1}x = y, then\)
\((a)\quad 0 \leq y \leq \pi \quad\quad (b)\quad -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \\
(c)\quad 0 < y < \pi \quad\quad (d)\quad -\frac{\pi}{2} < y < \frac{\pi}{2} \)
14. \(\tan^{-1}\sqrt{3} – \sec^{-1}(-2) \) is equal to
\((a) \pi \quad (b) -\frac{\pi}{3} \quad (c) \frac{\pi}{3} \quad (d) \frac{2\pi}{3}\)
Exercise 2.2
Prove the following:
\(\quad 1. \quad 3\sin^{-1}x = \sin^{-1}(3x – 4x^3), x \in \left [-\frac{1}{2}, \frac{1}{2}\right]\)
\(\quad 2. \quad 3\cos^{-1}x = \cos^{-1}(4x^3 – 3x), x \in \left [\frac{1}{2}, 1 \right]\)
Write the following functions in the simplest form:
\(\quad 3. \quad \tan^{-1} \frac{\sqrt{1 + x^2} – 1}{x}, x \neq 0 \)
\(\quad 4. \quad \tan^{-1} \left( \sqrt{\frac{1 – cosx}{1 + cosx}}\right), 0 < x < \pi\)
\(\quad 5. \quad \tan^{-1} \left(\frac{cos x – sinx}{cosx + sinx}\right), \frac{-\pi}{4} < x < \frac{3\pi}{4}\)
\(\quad 6. \quad \tan^{-1} \left(\frac{x}{\sqrt{a^2 – x^2}}\right), |x| < a \)
\(\quad 7. \quad \tan^{-1} \left(\frac{3a^2x – x^3}{a^3 – 3ax^2}\right), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}\)
Find the values of each of the following:
\(\quad 8. \quad \tan^{-1} \left[2cos \left(2 \sin^{-1} \frac{1}{2}\right) \right] \)
\(\quad 9. \quad \tan \frac{1}{2} \left[\sin^{-1}\frac{2x}{1 + x^2} + \cos^{-1} \frac{1 – y^2}{1 + y^2} \right] \)
Find the values of each of the expressions in Exercises 16 to 18.
\(\quad 10. \sin^{-1}\left(sin\frac{2\pi}{3}\right) \)
\(\quad 11. \tan^{-1}\left(tan\frac{3\pi}{4}\right) \)
\(\quad 12. \tan \left(sin^{-1}\frac{3}{5} + \cot^{-1} \frac{3}{2}\right) \)
\(\quad 13. \cos^{-1}\left(cos\frac{7\pi}{6}\right) \text{is equal to}\\
\quad (A) \frac{7\pi}{6} \quad (B) \frac{5\pi}{6} \quad (C) \frac{\pi}{3} \quad (D)\frac{\pi}{6}\)
\(\quad 14. \sin\left(\frac{\pi}{3} – sin^{-1}\left(-\frac{1}{2}\right) \right) \text{is equal to} \\
\quad (A) \frac{1}{2} \quad (B) \frac{1}{3} \quad (C) \frac{1}{4} \quad (D) 1 \)
\(\quad 15. \tan^{-1} – \cot^{-1}(-\sqrt{3}) \text{is equal to } \\
\quad (A) \pi \quad (B) -\frac{\pi}{2} \quad (C) 0 \quad (D) 2\sqrt{3}\)
Miscellaneous Exercise on Chapter 2
Find the value of the following:
\(\quad 1. \;\cos^{-1}\left(\cos\frac{13\pi}{6}\right) \)
\(\quad 2. \;\tan^{-1}\left(tan\frac{7\pi}{6}\right) \)
Prove that
\(\quad 3. \;2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}\)
\(\quad 4. \;\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{17}{36}\)
\(\quad 5. \;\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{13}{13}= \cos^{-1}\frac{33}{65}\)
\(\quad 6. \;\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5}= \sin^{-1}\frac{56}{65}\)
\(\quad 7. \;\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}\)
Prove that
\(\quad 8. \; \tan^{-1}\sqrt{x} = \frac{1}{2} \cos^{-1}\frac{1 – x}{1 + x}, x \in [0, 1]\)
\(\quad 9. \;\cot^{-1}\left(\frac{\sqrt{1 + sinx} + \sqrt{1 – sinx}}{\sqrt{1 + sinx} – \sqrt{1 – sinx}}\right)= \frac{x}{2}, x \in \left(0, \frac{\pi}{4}\right)\)
\(\quad 10. \;\tan^{-1}\left(\frac{\sqrt{1 + x} – \sqrt{1 – x}}{\sqrt{1 + x} + \sqrt{1 – x}}\right)= \frac{\pi}{4} – \frac{1}{2} \cos^{-1}x, -\frac{1}{\sqrt{2}} \leq x \leq 1 \)
Solve the following equations:
\(\quad 11. \;2\tan^{-1}(cosx) = \tan^{-1} (2cscx) \)
\(\quad 12. \; \tan^{-1}\frac{1 – x}{1 + x} = \frac{1}{2}\tan^{-1}x, (x > 0)\)
\(\quad 13. \; sin(\tan^{-1}x), |x| < 1 \,\text{is equal to}\\
\quad (A) \frac{x}{\sqrt{1 – x^2}} \quad (B) \frac{1}{\sqrt{1 – x^2}} \quad (C) \frac{1}{\sqrt{1 + x^2}} \quad (D) \frac{x}{\sqrt{1 + x^2}}\)
\(\quad 14. \; sin^{-1}(1 – x) – 2\sin^{-1}x = \frac{\pi}{2}, \text{then x is equal to}\\
\quad (A) 0, \frac{1}{2} \quad (B) 1, \frac{1}{2} \quad (C) 0 \quad (D) \frac{1}{2}\)
