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CBSE • Class XII • Mathematics • Ch 2
Estimated Time: 45 Mins
Study Progress: In Progress

Inverse Trigonometric Functions

In Class 12 Mathematics, "Inverse Trigonometric Functions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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📐 Have You Ever Wondered?

How do robotic arms calculate the exact joint angles needed to pick up a delicate glass wafer from Cartesian coordinates $(x, y, z)$? Inverse kinematics relies completely on restricted Principal Value Branches of inverse trigonometric functions.

Why This Chapter Matters

In Class 12 Mathematics, "Inverse Trigonometric Functions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Trigonometric functions from Class 11.
  • Bijective functions from Chapter 1.
  • Unit circle.

What You Will Learn (Core Objectives)

  • Define Inverse Trigonometric Functions and understand domain restriction for bijectivity.
  • State the Principal Value Branches (PVB) for $\sin^{-1} x, \cos^{-1} x, \tan^{-1} x, \text{cosec}^{-1} x, \sec^{-1} x, \cot^{-1} x$.
  • Evaluate principal values of inverse trigonometric expressions.
  • Apply negative argument identities: $\sin^{-1}(-x) = -\sin^{-1} x$ and $\cos^{-1}(-x) = \pi - \cos^{-1} x$.
  • Apply complementary sum identities: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.

Chapter Roadmap & Progression

1 1. Domain Restrictions & Principal...
2 2. Negative Argument & Complementar...

Complete Concept Guide (100% Curriculum Coverage)

1. Domain Restrictions & Principal Value Branches

Trigonometric functions are periodic and many-one, hence NOT invertible over $\mathbb{R}$. By restricting their domains to monotonic intervals, they become bijective with unique inverses:
• $\sin^{-1} x$: Domain $[-1, 1]$, Range $\mathbf{\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]}$.
• $\cos^{-1} x$: Domain $[-1, 1]$, Range $\mathbf{[0, \pi]}$.
• $\tan^{-1} x$: Domain $\mathbb{R}$, Range $\mathbf{\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)}$.

2. Negative Argument & Complementary Identities

  • Odd functions: $\sin^{-1}(-x) = -\sin^{-1} x, \quad \tan^{-1}(-x) = -\tan^{-1} x, \quad \text{cosec}^{-1}(-x) = -\text{cosec}^{-1} x$.
  • Supplementary shift: $\mathbf{\cos^{-1}(-x) = \pi - \cos^{-1} x}, \quad \sec^{-1}(-x) = \pi - \sec^{-1} x, \quad \cot^{-1}(-x) = \pi - \cot^{-1} x$.
  • Complementary Sums: $$\mathbf{\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}}, \quad \mathbf{\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}} \quad (|x| \le 1)$$

Inverse Trigonometric Functions - Key Conceptual & Analytical Model

Inverse Trigonometric Functions - Mathematical Architecture Axiomatic & Matrix Foundations Equivalence theorems & algebraic proofs Calculus & 3D Vector Geometry Differential optimization & spatial lines High-Stakes Examination & Engineering Mastery CBSE Class 12 Board criteria, JEE Advanced problem frameworks & applications

Chapter Summary & 10 Key Takeaways

Takeaway 1
Principal Value Branch: Canonical range interval ensuring unique single-valued function output.
Takeaway 2
Domain Boundaries: Restricted input intervals ($[-1, 1]$ for sine and cosine inverses).
Takeaway 3
Cosine Negative Rule: $\cos^{-1}(-x) = \pi - \cos^{-1} x$ mapping strictly into Quadrant II $[0, \pi]$.
Takeaway 4
Complementary Orthogonality: Sum of co-functions equals $\pi/2$ (90 degrees).
Takeaway 5
Self-Inversion Limit: $\sin^{-1}(\sin x) = x$ ONLY if $x \in [-\pi/2, \pi/2]$.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Find the principal value of $\cos^{-1}\left(-\frac{1}{2}\right)$.
Reveal Answer & Explanation
Answer: Let $y = \cos^{-1}\left(-\frac{1}{2}\right) \implies \cos y = -\frac{1}{2} = -\cos\frac{\pi}{3} = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\frac{2\pi}{3}$. Since $\frac{2\pi}{3} \in [0, \pi]$ (the PVB of $\cos^{-1}$), the principal value is $\frac{2\pi}{3}$.
2π / 3.
2
Evaluate: $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$.
Reveal Answer & Explanation
Answer: The range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$. Since $\frac{2\pi}{3} \notin [-\pi/2, \pi/2]$, we cannot write $\frac{2\pi}{3}$ directly. Rewrite: $\sin\frac{2\pi}{3} = \sin(\pi - \frac{\pi}{3}) = \sin\frac{\pi}{3}$. Therefore, $\sin^{-1}\left(\sin\frac{\pi}{3}\right) = \frac{\pi}{3} \in [-\pi/2, \pi/2]$.
π / 3.
3
Find the value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$.
Reveal Answer & Explanation
Answer: $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$. $\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$. Expression $= \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$.
-π / 3.
4
State the domain and range of the principal value branch of $f(x) = \sin^{-1} x$.
Reveal Answer & Explanation
Answer: Domain is $[-1, 1]$ and Range (Principal Value Branch) is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Domain: [-1, 1], Range: [-π/2, π/2].
5
Find the value of $x$ if $\sin(\sin^{-1}\frac{1}{5} + \cos^{-1} x) = 1$.
Reveal Answer & Explanation
Answer: $\sin^{-1}\frac{1}{5} + \cos^{-1} x = \sin^{-1}(1) = \frac{\pi}{2}$. Since $\sin^{-1} a + \cos^{-1} a = \frac{\pi}{2}$, we have $x = \frac{1}{5}$.
x = 1/5.
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