The principal value of \(\sec^{-1}\left(-\dfrac{2}{\sqrt3}\right)\) is
- (a)\(-\dfrac{\pi}{6}\)
- (b)\(\dfrac{5\pi}{6}\)
- (c)\(\dfrac{7\pi}{6}\)
- (d)\(\dfrac{2\pi}{3}\)
Inverse Trigonometric Functions Class 12: notes and important questions
Revision notes, 39 board-style questions with the step marks shown, and a four-step route from the basics to 95+, each step ending in a short checkpoint.
- 39 questions
- 12 multiple choice, 3 assertion–reason, 8 very short answer, 12 short answer, 1 long answer, 3 case study
- About 10 hours to master
Relations and Functions unit: 8 of 80 theory marks (Relations and Functions and Inverse Trigonometric Functions).
Revision notes
Inverse Trigonometric Functions — revision notes
1. Principal value branches
| Function | Domain | Range (principal branch) |
|---|---|---|
| \(\sin^{-1}x\) | \([-1, 1]\) | \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) |
| \(\cos^{-1}x\) | \([-1, 1]\) | \([0, \pi]\) |
| \(\tan^{-1}x\) | \(\mathbb{R}\) | \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) |
| \(\text{cosec}^{-1}x\) | \(\mathbb{R} - (-1, 1)\) | \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\) |
| \(\sec^{-1}x\) | \(\mathbb{R} - (-1, 1)\) | \([0, \pi] - \left\{\frac{\pi}{2}\right\}\) |
| \(\cot^{-1}x\) | \(\mathbb{R}\) | \((0, \pi)\) |
\(\sin^{-1}x\) means “the angle in the principal range whose sine is \(x\)”. It is not \(\dfrac{1}{\sin x}\).
2. Graphs
The graph of an inverse function is the mirror image of the (restricted) original graph in the line \(y = x\). \(\sin^{-1}\) and \(\tan^{-1}\) are increasing and their graphs are symmetric about the origin (read this from the graph; the identity \(\sin^{-1}(-x) = -\sin^{-1}x\) belongs to the deleted properties section); \(\cos^{-1}\) is decreasing from \(\pi\) to \(0\); \(\tan^{-1}\) has horizontal asymptotes \(y = \pm\dfrac{\pi}{2}\).
3. Compositions
- \(\sin(\sin^{-1}x) = x\) for \(x \in [-1, 1]\) (similarly for the others on their domains).
- \(\sin^{-1}(\sin\theta) = \theta\) only for \(\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\). Otherwise first rewrite \(\sin\theta = \sin\phi\) with \(\phi\) in the principal range (using \(\sin(\pi - \theta) = \sin\theta\) and period \(2\pi\)); the answer is \(\phi\).
- Same idea for \(\cos^{-1}(\cos\theta)\) (use \(\cos(2\pi - \theta) = \cos\theta\)) and \(\tan^{-1}(\tan\theta)\) (period \(\pi\)).
- For \(\tan\left(\cos^{-1}\frac35\right)\)-type questions, draw a right triangle, but check the sign using the principal range.
Worked example 1
\(\sin^{-1}\left(-\dfrac12\right) = -\dfrac{\pi}{6}\), \(\cos^{-1}\left(-\dfrac12\right) = \dfrac{2\pi}{3}\).
Worked example 2
\(\tan^{-1}\left(\tan\dfrac{5\pi}{6}\right)\): \(\tan\dfrac{5\pi}{6} = \tan\left(\dfrac{5\pi}{6} - \pi\right) = \tan\left(-\dfrac{\pi}{6}\right)\), so the value is \(-\dfrac{\pi}{6}\).
Worked example 3
Domain of \(\sin^{-1}(2x + 1)\): \(-1 \le 2x + 1 \le 1 \Rightarrow -1 \le x \le 0\).
Common errors
- Writing \(\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3}\) — the answer must lie in the principal range.
- Giving \(\cos^{-1}(-x)\) a negative value: \(\cos^{-1}\) is never negative.
- Mixing up ranges of \(\cot^{-1}\) \((0, \pi)\) and \(\tan^{-1}\) \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
- Ignoring the sign of a square root after a substitution (e.g. \(\sqrt{\cos^2\theta} = |\cos\theta|\)).
Board-exam tips
- State the principal range you are using in each step — it is where the marks are.
- The general property formulas (sum of \(\tan^{-1}\), \(\sin^{-1}x + \cos^{-1}x\), etc.) are no longer in the syllabus; reason from the branches instead.
- Values in radians: \(\sin 2\) means \(2\) radians.
Topics in this chapter: Principal values · Domain and range (principal value branches) · Compositions and simplification · Graphs of inverse trigonometric functions.
Route to 95: four steps
Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.
Step 1
Secure the basics
You can write the principal value branch of every inverse trig function and find principal values of standard values.
Step 2
Board standard
You can find domains, simplify compositions like sin⁻¹(sin x) using the principal range and evaluate expressions such as cos(tan⁻¹ ¾).
Step 3
Full marks on long answers
You can write complete multi-part answers (the case studies, the 5-mark simplification and the multi-part domain questions), stating the branch you use at each step.
Step 4
95+ stretch (HOTS)
You can handle radian inputs like sin⁻¹(sin 3), substitution-based simplifications and justifying the signs when you write sin θ and cos θ in terms of x.
Practice questions
Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 3 of the 39 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.
If \(\sin^{-1}(2x - 1) = \dfrac{\pi}{6}\), find \(x\) and hence the value of \(\cos^{-1}(1 - 2x)\).
The range of the principal value branch of \(\sec^{-1}x\) is
- (a)\(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\)
- (b)\([0, \pi] - \left\{\dfrac{\pi}{2}\right\}\)
- (c)\((0, \pi)\)
- (d)\(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}\)
Where marks are lost in Inverse Trigonometric Functions
- Writing sin⁻¹(sin 2π/3) = 2π/3. Fix: rewrite the angle into the principal range first (sin 2π/3 = sin π/3) and state the range.
- Negative values for cos⁻¹ or cot⁻¹. Fix: cos⁻¹ lies in [0, π] and cot⁻¹ in (0, π); if you get a negative answer, re-check.
- Mixing the ranges of tan⁻¹ and cot⁻¹. Fix: write the branch table at the top of your answer for any mixed question.
- Dropping the modulus after a substitution (√cos²θ = |cos θ|). Fix: state the interval of θ and remove the modulus with the correct sign.
- Using the deleted property formulas (tan⁻¹x + tan⁻¹y, etc.). Fix: they are out of the syllabus; reason from the principal branches and right-triangle values instead.
- Treating sin 2 as sin 2°. Fix: numbers without a degree sign are radians; use 3 < π < 3.2, so π/2 is just over 1.5 and π is just over 3 (no calculator needed).
Inverse Trigonometric Functions in our sample papers
Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).
- Sample paper 1 (Board standard): questions 2, 21
- Sample paper 2 (Board standard): questions 2, 21
- Sample paper 3 (Above board): questions 2, 21
- Sample paper 4 (Above board): questions 2, 21
- Sample paper 5 (95+ challenge): questions 2, 21