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MathematicsClass 12Mathematics

Mathematics | Class 12 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 2 min read

Mathematics – this guide gives you a concise, exam-ready overview of Mathematics from Class 12 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

2.1 Introduction

This section introduces the concept of inverse functions in the context of trigonometric functions. It recalls from Chapter 1 that a function f has an inverse f⁻¹ if and only if f is one-one (injective) and onto (surjective). Many functions, including the standard trigonometric functions, are not one-one and onto over their natural domains and ranges, so their inverses do not exist without domain restrictions. The section explains that by restricting the domain and range of trigonometric functions appropriately, we can define their inverses. These inverse trigonometric functions are crucial in calculus, especially in defining integrals, and have wide applications in science and engineering. The section also pays homage to Aryabhata, an ancient Indian mathematician who contributed to trigonometry. The importance of inverse trigonometric functions is emphasized as foundational for further mathematical study.

📊 Diagram: Image of Aryabhata, an ancient Indian mathematician (476-550 A.D.), highlighting the historical roots of trigonometry in India.

🧪 Activity: No specific activity in this section.

🔗 Connection: Leads to the detailed study of domain and range restrictions for trigonometric functions to define their inverses in section 2.2.

Frequently asked questions

Find the principal values of the following: 1. \(\sin^{-1}\left(-\frac{1}{2}\right)\) 2. \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\) 3. \(\csc^{-1}(2)\) 4. \(\tan^{-1}(-\sqrt{3})\) 5. \(\cos^{-1}\left(-\frac{1}{2}\right)\) 6. \(\tan^{-1}(-1)\) 7. \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\) 8. \(\cot^{-1}(\sqrt{3})\) 9. \(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) 10. \(\csc^{-1}(-\sqrt{2})\)

1. \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\) because principal value of \(\sin^{-1}x\) lies in \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

2. \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}\) since principal value of \(\cos^{-1}x\) lies in \([0, \pi]\).

3. \(\csc^{-1}(2) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).

4. \(\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}\) as principal value of \(\tan^{-1}x\) lies in \((-\frac{\pi}{2}, \frac{\pi}{2})\).

5. \(\cos^{-1}\left(-\fr

Find the values of the following: 11. \(\tan^{-1}(1) + \cos^{-1} - \frac{1}{2} + \sin^{-1} - \frac{1}{2}\) 12. \(\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2}\)

11. Evaluate each term:

  • \(\tan^{-1}(1) = \frac{\pi}{4}\)
  • \(\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\)
  • \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\)

Sum = \(\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}\).

12. \(\cos^{-1}\frac{1}{2} = \frac{\pi}{3}\), \(\sin^{-1}\frac{1}{2} = \frac{\pi}{6}\)

Therefore, \(\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2} = \frac{\pi}{3} + 2 \times \fr

13. If \(\sin^{-1}x = y\) , then (A) \(0 \leq y \leq \pi\) (B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\) (C) \(0 < y < \pi\) (D) \(-\frac{\pi}{2} < y < \frac{\pi}{2}\)

\(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)

Explanation: The principal value branch of \(\sin^{-1}x\) is defined such that \(y = \sin^{-1}x\) lies in \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

14. \(\tan^{-1}\sqrt{3} -\sec^{-1}(-2)\) is equal to (A) \(\pi\) (B) \(-\frac{\pi}{3}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{2\pi}{3}\)

\(\frac{2\pi}{3}\)

Explanation:

  • \(\tan^{-1}\sqrt{3} = \frac{\pi}{3}\)
  • \(\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\)

Therefore, \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\).

But since the options include \(-\frac{\pi}{3}\) as (B), check carefully: Actually, the question asks for the value, so the answer is (B) \(-\frac{\pi}{3}\).

Hence correct answer is (B) \(-\frac{\pi}{3}\).

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