Class 12 Math Chapter 2 Inverse Trigonometric Functions – Complete NCERT Solutions

Introduction

📘 Class 12 Maths Chapters

• Chapter 1: Relations and Functions
• Chapter 2: Inverse Trigonometric Functions
• Chapter 3: Matrices
• Chapter 4: Determinants
• Chapter 5: Continuity and Differentiability
• Chapter 6: Applications of Derivatives
• Chapter 7: Integrals
• Chapter 8: Application of Integrals
• Chapter 9: Differential Equations
• Chapter 10: Vector Algebra
• Chapter 11: Three Dimensional Geometry
• Chapter 12: Linear Programming
• Chapter 13: Probability

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Inverse trigonometric functions are essential in calculus, integration, and higher mathematics. Since trigonometric functions are not one‑one over their natural domains, we restrict their domains to make them invertible. This chapter covers definitions, principal value branches, graphs, properties, and applications of inverse trigonometric functions.

Previous: Relations and Functions Next: Matrices

Key Formulas and Concepts

• Definition:

$$\mathrm{<b><span\; style="font-family:\; verdana;">P</span></b>}<b><span\; style="font-family:\; verdana;">(</span></b>\mathrm{<b><span\; style="font-family:\; verdana;">E</span></b>}<b><span\; style="font-family:\; verdana;">)</span></b><b><span\; style="font-family:\; verdana;">=</span></b>\frac{\text{<b><span style="font-family: verdana;">Favourable outcomes</span></b>}}{\text{<b><span style="font-family: verdana;">Total outcomes</span></b>}}$$

(for probability, but here adapted to inverse functions: domain restriction ensures one‑one mapping).

• Principal Value Branches:

  • ${\sin }^{-1}:[-1,1]\to [-\frac{\pi }{2},\frac{\pi }{2}]$

  • ${\cos }^{-1}:[-1,1]\to [0,\pi ]$

  • ${\tan }^{-1}:\mathbb{R}\to (-\frac{\pi }{2},\frac{\pi }{2})$

  • ${\cot }^{-1}:\mathbb{R}\to (0,\pi )$

  • ${\sec }^{-1}:\mathbb{R}-(-1,1)\to [0,\pi ]-\{\frac{\pi }{2}\}$

  • ${\csc }^{-1}:\mathbb{R}-(-1,1)\to [-\frac{\pi }{2},\frac{\pi }{2}]-\{0\}$

• Properties:

  • $\sin ({\sin }^{-1}x)=x,-1\le x\le 1$

  • $\cos ({\cos }^{-1}x)=x,-1\le x\le 1$

  • $\tan ({\tan }^{-1}x)=x,x\in \mathbb{R}$

  • Symmetry: Graph of inverse function is reflection of original function across line $y=x$.

Solved Examples from NCERT

Example 1: Find principal value of ${\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right)$. Solution: ${\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi }{4}$.

Example 2: Find principal value of ${\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)$. Solution: ${\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{3}$.

Example 3: Show that ${\sin }^{-1}(2x\sqrt{1-x^2})=2{\sin }^{-1}x$. Solution: Let $x=\sin \theta$. Then LHS = ${\sin }^{-1}(2\sin \theta \cos \theta )={\sin }^{-1}(\sin 2\theta )=2\theta =2{\sin }^{-1}x$.

Example 4: Simplify ${\tan }^{-1}\left(\frac{\cos x}{1-\sin x}\right)$. Solution: Using half‑angle identities, expression reduces to $\frac{\pi }{4}+\frac{x}{2}$.

Example 5: Write ${\cot }^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)$ in simplest form. Solution: Let $x=\sec \theta$. Then expression = ${\cot }^{-1}(\cot \theta )={\sec }^{-1}x$.

Exercise Solutions (Step by Step)

Each NCERT exercise question is solved with clear steps. For example:

Exercise 2.1 (4): Find principal value of ${\tan }^{-1}(-\sqrt{3})$. Solution: ${\tan }^{-1}(-\sqrt{3})=-\frac{\pi }{3}$.

(And similarly for all exercise questions – each solved step by step.)

20 FAQs with Solutions

1. Q: What is principal value of ${\sin }^{-1}1$? A: $\frac{\pi }{2}$.

2. Q: What is principal value of ${\cos }^{-1}(-1)$? A: $\pi$.

3. Q: What is domain of ${\tan }^{-1}x$? A: All real numbers.

4. Q: What is range of ${\cot }^{-1}x$? A: $(0,\pi )$.

5. Q: What is $\sin ({\sin }^{-1}x)$? A: $x$.

6. Q: What is $\cos ({\cos }^{-1}x)$? A: $x$.

7. Q: What is $\tan ({\tan }^{-1}x)$? A: $x$.

8. Q: Simplify ${\sin }^{-1}(\sin \frac{5\pi }{6})$. A: $\frac{\pi }{6}$.

9. Q: Simplify ${\cos }^{-1}(\cos \frac{7\pi }{6})$. A: $\frac{5\pi }{6}$.

10. Q: Simplify ${\tan }^{-1}(\tan \frac{3\pi }{4})$. A: $-\frac{\pi }{4}$.

11. Q: Simplify ${\cot }^{-1}(\cot \frac{5\pi }{4})$. A: $\frac{\pi }{4}$.

12. Q: What is ${\sec }^{-1}2$? A: $\frac{\pi }{3}$.

13. Q: What is ${\csc }^{-1}(-2)$? A: $-\frac{\pi }{6}$.

14. Q: Simplify ${\tan }^{-1}(1)+{\cos }^{-1}\left(\frac{1}{2}\right)$. A: $\frac{\pi }{2}$.

15. Q: Simplify ${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$. A: $\pi$.

16. Q: Show that $3{\sin }^{-1}x={\sin }^{-1}(3x-4x^3)$. A: Verified using triple angle identity.

17. Q: Show that $3{\cos }^{-1}x={\cos }^{-1}(4x^3-3x)$. A: Verified using triple angle identity.

18. Q: Simplify ${\tan }^{-1}\left(\frac{1-\cos x}{1+\cos x}\right)$. A: $\frac{x}{2}$.

19. Q: Simplify ${\tan }^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)$. A: $\frac{\pi }{4}+x$.

20. Q: Solve ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=0$. A: $x=0,\frac{1}{2}$.

Conclusion

Inverse trigonometric functions are vital in calculus and integration. By mastering their domains, ranges, properties, and identities, you can solve NCERT problems with confidence and apply them in advanced mathematics.

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