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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.

2.3 Properties of Inverse Trigonometric Functions

This section discusses important properties of inverse trigonometric functions, valid within their principal value branches and domains. It recalls that if y = sin⁻¹ x, then x = sin y, and vice versa, with sin(sin⁻¹ x) = x for x in [-1, 1] and sin⁻¹(sin x) = x for x in [-π/2, π/2]. Similar properties hold for other inverse trig functions within their domains. The section emphasizes that some properties may not hold outside these principal ranges. It includes proofs of identities such as sin⁻¹(2x√(1 - x²)) = 2 sin⁻¹ x for x in [-1/√2, 1/√2], and sin⁻¹(2x√(1 - x²)) = 2 cos⁻¹ x for x in [1/√2, 1]. These are proved by substituting x = sin θ or x = cos θ and using double-angle formulas. Another example simplifies tan⁻¹(cos x / (1 - sin x)) to (π/4) + (x/2) using half-angle identities and the addition formula for tangent. The section also includes an example expressing cot⁻¹(1/√(x² - 1)) as sec⁻¹ x for x > 1. These properties and examples are foundational for solving complex trigonometric expressions and integrals.

📊 Diagram: No diagrams; focuses on algebraic proofs and simplifications.

🧪 Activity: Exercise 2.2: Prove identities and simplify inverse trigonometric expressions using properties.

🔗 Connection: Prepares for miscellaneous examples and exercises applying these properties in the next section.

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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