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MathematicsClass 12Mathematics

Mathematics | Class 12 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 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.

Historical Note

This section provides a historical overview of the development of trigonometry, emphasizing its origins in ancient India. It credits Indian mathematicians such as Aryabhata (476 A.D.), Brahmagupta (598 A.D.), Bhaskara I (600 A.D.), and Bhaskara II (1114 A.D.) for significant contributions to trigonometry, including the development of sine functions and formulas for sine values of angles greater than 90°. The knowledge spread from India to Arabia and then to Europe, where it was adopted due to its superior approach compared to earlier Greek methods. The section mentions the use of sine in Sanskrit astronomical works (siddhantas) and highlights Bhaskara I's formulae and Bhaskara II's exact expressions for sines of special angles. It notes that the notation sin⁻¹, cos⁻¹, etc., was introduced by Sir John F.W. Herschel in 1813. The section also references Thales of Greece (about 600 B.C.) and his contributions to height and distance problems using trigonometric ratios, illustrating the practical applications of trigonometry in ancient times. This historical context enriches the understanding of the subject's development and its global significance.

📊 Diagram: No diagrams; historical images or illustrations of ancient mathematicians could be referenced.

🧪 Activity: No activity; serves as historical context.

🔗 Connection: Provides background enriching the understanding of the mathematical concepts studied in the chapter.

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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