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Chapter 2 · Inverse Trigonometric Functions · Go for it

Inverse trig functions, the quick way

Every answer in this chapter is an angle from a fixed range. Learn the six ranges and seven moves, and you can answer most questions.

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The chapter in one minute

What an inverse trig function is, and what the exam asks.

sin⁡−1x\sin^{-1}x asks a question: which angle has sine equal to xx⁠? Its answer is an angle.

Many angles have the same sine, so each inverse picks its answer from one fixed range. This choice is called the principal value branch, and the answer from that range is the principal value.

So the whole chapter rests on six ranges. Learn them by heart, and always check that your answer lies in the right one.

Most exam questions on this chapter are one of seven kinds, and each kind has a fixed recipe. You will learn all seven below.

The book also has "simplest form" questions and equations to solve. They are in the book, but the CBSE syllabus leaves them out, so they are not on this page.

7question types
12formula cards
6ranges to learn by heart
1check that catches most mistakes: is the answer inside the range?
Formulas

The formulas you must know

The first six cards are the ranges, exactly as the book gives them. Learn these cards first.

The range of sin⁡−1\sin^{-1}
sin⁡−1:[−1,1]→[−π2,π2]\sin^{-1} : [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Answers from −π2-\frac{\pi}{2} to π2\frac{\pi}{2}⁠, both ends included. A negative input gives a negative answer.

The range of cos⁡−1\cos^{-1}
cos⁡−1:[−1,1]→[0,π]\cos^{-1} : [-1, 1] \to [0, \pi]

Answers from 0 to π\pi⁠. Never negative.

The range of tan⁡−1\tan^{-1}
tan⁡−1:R→(−π2,π2)\tan^{-1} : \R \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Round brackets: the ends are left out, because tan⁡π2\tan\frac{\pi}{2} does not exist.

The range of cot⁡−1\cot^{-1}
cot⁡−1:R→(0,π)\cot^{-1} : \R \to (0, \pi)

Never negative, like cos⁡−1\cos^{-1}⁠, but 0 and π\pi are left out.

The range of sec⁡−1\sec^{-1}
sec⁡−1:R−(−1,1)→[0,π]−{π2}\begin{aligned} &\sec^{-1} : \R - (-1, 1) \\ &\to [0, \pi] - \left\{\frac{\pi}{2}\right\} \end{aligned}

The range of cos⁡−1\cos^{-1} without π2\frac{\pi}{2}⁠.

The range of cosec⁡−1\operatorname{cosec}^{-1}
cosec⁡−1:R−(−1,1)→[−π2,π2]−{0}\begin{aligned} &\operatorname{cosec}^{-1} : \R - (-1, 1) \\ &\to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\} \end{aligned}

The range of sin⁡−1\sin^{-1} without 0.

Special values of sin and cos
sin⁡π6=cos⁡π3=12sin⁡π4=cos⁡π4=12sin⁡π3=cos⁡π6=32\begin{aligned} \sin\frac{\pi}{6} &= \cos\frac{\pi}{3} = \frac12 \\ \sin\frac{\pi}{4} &= \cos\frac{\pi}{4} = \frac{1}{\sqrt2} \\ \sin\frac{\pi}{3} &= \cos\frac{\pi}{6} = \frac{\sqrt3}{2} \end{aligned}

Also sin⁡π2=1\sin\frac{\pi}{2} = 1 and cos⁡0=1\cos 0 = 1⁠.

Special values of tan
tan⁡π6=13tan⁡π4=1tan⁡π3=3\begin{aligned} \tan\frac{\pi}{6} &= \frac{1}{\sqrt3} \\ \tan\frac{\pi}{4} &= 1 \\ \tan\frac{\pi}{3} &= \sqrt3 \end{aligned}

cot is 1 over tan: cot⁡π6=3\cot\frac{\pi}{6} = \sqrt3⁠.

Undoing
sin⁡−1(sin⁡x)=x,x∈[−π2,π2]\begin{aligned} &\sin^{-1}(\sin x) = x, \\ &x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}

Only for these xx⁠. In the same way, cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x for x∈[0,π]x \in [0, \pi]⁠, and tan⁡−1(tan⁡x)=x\tan^{-1}(\tan x) = x for x∈(−π2,π2)x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠.

Moving an angle
sin⁡(π−θ)=sin⁡θcos⁡(π−θ)=−cos⁡θtan⁡(π+θ)=tan⁡θ\begin{aligned} \sin(\pi - \theta) &= \sin\theta \\ \cos(\pi - \theta) &= -\cos\theta \\ \tan(\pi + \theta) &= \tan\theta \end{aligned}

Also cos⁡(2π−θ)=cos⁡θ\cos(2\pi - \theta) = \cos\theta and cos⁡(2π+θ)=cos⁡θ\cos(2\pi + \theta) = \cos\theta⁠. Use these to bring an angle into the range.

The triangle trick
cos⁡α=1−sin⁡2α\cos\alpha = \sqrt{1 - \sin^2\alpha}

True for α\alpha in the range of sin⁡−1\sin^{-1}⁠. For sin⁡−135\sin^{-1}\frac35⁠, draw a right triangle with sides 3, 4, 5.

Tan of a sum
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\begin{aligned} &\tan(A + B) \\ &= \frac{\tan A + \tan B}{1 - \tan A\tan B} \end{aligned}

Find tan⁡A\tan A and tan⁡B\tan B first, then put them in.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Principal value of a positive number: find the angle from 0 to π2\frac{\pi}{2}

Spot it: an inverse of a positive number, such as cos⁡−132\cos^{-1}\frac{\sqrt3}{2} or cosec⁡−12\operatorname{cosec}^{-1}2⁠.

  1. Call the answer yy⁠, and write the matching sin, cos or tan of yy⁠.
  2. Find the angle on the special-values cards. For sec⁡−1\sec^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠, turn the number upside down and use cos or sin.
  3. Write the range line, then the answer in radians.

Exercise 2.1 Q2. Find the principal value of cos⁡−1(32)\cos^{-1}\left(\frac{\sqrt3}{2}\right)⁠.

Let cos⁡−1(32)=y\cos^{-1}\left(\frac{\sqrt3}{2}\right) = y⁠. Then

cos⁡y=32=cos⁡π6\cos y = \frac{\sqrt3}{2} = \cos\frac{\pi}{6}

Range of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠, and π6∈[0,π]\frac{\pi}{6} \in [0, \pi]⁠.

∴ cos⁡−1(32)=π6\cos^{-1}\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{6}

See a full exam answer · Go deeper: see it visually · Basics: special angles

2. A negative number in sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠: add a minus sign

Spot it: a negative number inside sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠.

  1. Ignore the minus sign and find the angle for the positive number, as in type 1.
  2. Put the minus sign on the angle. These three ranges go down to −π2-\frac{\pi}{2}⁠, so a negative answer is allowed.
  3. Write the range line, then the answer.

Exercise 2.1 Q4. Find the principal value of tan⁡−1(−3)\tan^{-1}\left(-\sqrt3\right)⁠.

Let tan⁡−1(−3)=y\tan^{-1}\left(-\sqrt3\right) = y⁠. Then

tan⁡y=−3=−tan⁡π3=tan⁡(−π3)\begin{aligned} \tan y &= -\sqrt3 = -\tan\frac{\pi}{3} \\ &= \tan\left(-\frac{\pi}{3}\right) \end{aligned}

Range of tan⁡−1\tan^{-1} is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, and −π3∈(−π2,π2)-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠.

∴ tan⁡−1(−3)=−π3\tan^{-1}\left(-\sqrt3\right) = -\frac{\pi}{3}

See a full exam answer · Go deeper: see it visually · Basics: sin and cos on the unit circle

In past papers (2): click to see
Board 2024 Q38(ii) · Sample 2025-26 Q21(a)

3. A negative number in cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠: take it away from π\pi

Spot it: a negative number inside cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠.

  1. Ignore the minus sign and find the angle for the positive number, as in type 1.
  2. Take it away from π\pi⁠. These ranges start at 0, so the answer is never negative.
  3. Write the range line, then the answer.

Exercise 2.1 Q5. Find the principal value of cos⁡−1(−12)\cos^{-1}\left(-\frac12\right)⁠.

Let cos⁡−1(−12)=y\cos^{-1}\left(-\frac12\right) = y⁠. Using cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos\theta⁠,

cos⁡y=−12=−cos⁡π3=cos⁡(π−π3)=cos⁡2π3\begin{aligned} \cos y &= -\frac12 = -\cos\frac{\pi}{3} \\ &= \cos\left(\pi - \frac{\pi}{3}\right) \\ &= \cos\frac{2\pi}{3} \end{aligned}

Range of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠, and 2π3∈[0,π]\frac{2\pi}{3} \in [0, \pi]⁠.

∴ cos⁡−1(−12)=2π3\cos^{-1}\left(-\frac12\right) = \frac{2\pi}{3}

See a full exam answer · Go deeper: see it visually · Basics: sin and cos on the unit circle

In past papers (2): click to see
Board 2026 Q25(b) · Sample 2022-23 Q19

4. Adding principal values: one at a time, then add

Spot it: two or three inverses joined by plus or minus signs, and the question asks for the value.

  1. Find each principal value on its own line, with its range.
  2. Put the values in and add, using a common denominator for the fractions of π\pi⁠.

Exercise 2.1 Q12. Find the value of cos⁡−1(12)+2sin⁡−1(12)\cos^{-1}\left(\frac12\right) + 2\sin^{-1}\left(\frac12\right)⁠.

cos⁡−112=π3\cos^{-1}\frac12 = \frac{\pi}{3}⁠, as π3∈[0,π]\frac{\pi}{3} \in [0, \pi]⁠.

sin⁡−112=π6\sin^{-1}\frac12 = \frac{\pi}{6}⁠, as π6∈[−π2,π2]\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

cos⁡−112+2sin⁡−112=π3+2⋅π6=π3+π3\begin{aligned} &\cos^{-1}\frac12 + 2\sin^{-1}\frac12 \\ &= \frac{\pi}{3} + 2\cdot\frac{\pi}{6} \\ &= \frac{\pi}{3} + \frac{\pi}{3} \end{aligned}

∴ The value is 2π3\frac{2\pi}{3}

See a full exam answer · Go deeper: see it visually

In past papers (3): click to see
Board 2024 Q38(ii) · Board 2023 Q21(b) · Sample 2025-26 Q19

5. An inverse of its own trig function: move the angle into the range first

Spot it: sin⁡−1(sin⁡…)\sin^{-1}(\sin\ldots)⁠, cos⁡−1(cos⁡…)\cos^{-1}(\cos\ldots) or tan⁡−1(tan⁡…)\tan^{-1}(\tan\ldots) with a given angle.

  1. Is the angle inside the range? If yes, the answer is the angle itself.
  2. If not, change it to an angle inside the range with the same sin, cos or tan, using the moving card.
  3. Write the range line. Now the inverse undoes the trig function.

Exercise 2.2 Q10. Find the value of sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)⁠.

2π3∉[−π2,π2]\frac{2\pi}{3} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, the range of sin⁡−1\sin^{-1}⁠. Using sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta⁠,

sin⁡−1(sin⁡2π3)=sin⁡−1(sin⁡(π−π3))=sin⁡−1(sin⁡π3)\begin{aligned} &\sin^{-1}\left(\sin\frac{2\pi}{3}\right) \\ &= \sin^{-1}\left(\sin\left(\pi - \frac{\pi}{3}\right)\right) \\ &= \sin^{-1}\left(\sin\frac{\pi}{3}\right) \end{aligned}

As π3∈[−π2,π2]\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠,

∴ sin⁡−1(sin⁡2π3)=π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}

See a full exam answer · Go deeper: see it visually · Basics: reference angles

In past papers (3): click to see
Board 2023 Q21(b) · Sample 2023-24 Q21(a) · Sample 2022-23 Q21(a)

6. A trig function of an inverse: work from the inside out

Spot it: an inverse sitting inside sin, cos or tan, like cos⁡(2sin⁡−112)\cos\left(2\sin^{-1}\frac12\right)⁠.

  1. Find the innermost inverse first, as a principal value.
  2. Put it in and simplify, one bracket at a time.
  3. If an inverse is left on the outside, find its principal value too.

Exercise 2.2 Q8. Find the value of tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]⁠.

sin⁡−112=π6\sin^{-1}\frac12 = \frac{\pi}{6}⁠, as π6∈[−π2,π2]\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

tan⁡−1[2cos⁡(2sin⁡−112)]=tan⁡−1(2cos⁡π3)=tan⁡−1(2⋅12)=tan⁡−11\begin{aligned} &\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right] \\ &= \tan^{-1}\left(2\cos\frac{\pi}{3}\right) \\ &= \tan^{-1}\left(2\cdot\frac12\right) \\ &= \tan^{-1}1 \end{aligned}

As tan⁡π4=1\tan\frac{\pi}{4} = 1 and π4∈(−π2,π2)\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠,

∴ The value is π4\frac{\pi}{4}

See a full exam answer · Go deeper: see it visually

In past papers (4): click to see
Board 2026 Q25(b) · Board 2025 Q22 · Sample 2026-27 Q3 · Sample 2025-26 Q21(a)

7. Fractions like 35\frac35 inside the inverses: the triangle trick

Spot it: tan of a sum such as sin⁡−135+cot⁡−132\sin^{-1}\frac35 + \cot^{-1}\frac32⁠, or a "prove that" with fractions like 817\frac{8}{17} and 1213\frac{12}{13}⁠.

  1. Give each inverse a name, like α\alpha and β\beta⁠. A positive fraction like 35\frac35 gives an acute angle.
  2. For each one, use the triangle card (or draw the right triangle) and read off its tan.
  3. Put the two tans into the tan of a sum card and simplify.
  4. In a "prove that", check the angle lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) before you write it as tan⁡−1\tan^{-1} of the value.

Exercise 2.2 Q12. Find the value of tan⁡(sin⁡−135+cot⁡−132)\tan\left(\sin^{-1}\frac35 + \cot^{-1}\frac32\right)⁠.

Let α=sin⁡−135\alpha = \sin^{-1}\frac35 and β=cot⁡−132\beta = \cot^{-1}\frac32⁠. Both are acute.

sin⁡α=35\sin\alpha = \frac35⁠, so cos⁡α=45\cos\alpha = \frac45 and tan⁡α=34\tan\alpha = \frac34⁠.

cot⁡β=32\cot\beta = \frac32⁠, so tan⁡β=23\tan\beta = \frac23⁠.

tan⁡(α+β)=34+231−34⋅23=171212\begin{aligned} &\tan(\alpha + \beta) \\ &= \frac{\frac34 + \frac23}{1 - \frac34\cdot\frac23} \\ &= \frac{\frac{17}{12}}{\frac12} \end{aligned}

∴ The value is 176\frac{17}{6}

See a full exam answer · Go deeper: see it visually

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

  • Learn the six range cards first. Half the chapter is just these.
  • Before you write an answer, ask: is it inside the range? It takes five seconds.
  • Negative number in sin⁡−1\sin^{-1}⁠, tan⁡−1\tan^{-1} or cosec⁡−1\operatorname{cosec}^{-1}⁠? Just add a minus sign: sin⁡−1(−12)=−π6\sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}⁠.
  • Negative number in cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠? Take the angle for the positive number away from π\pi⁠: cos⁡−1(−12)=π−π3\cos^{-1}\left(-\frac12\right) = \pi - \frac{\pi}{3}⁠.
  • sec⁡−1\sec^{-1} and cosec⁡−1\operatorname{cosec}^{-1}⁠: turn the number upside down and use cos or sin, for example sec⁡−12=cos⁡−112\sec^{-1}2 = \cos^{-1}\frac12⁠.
  • Give answers in radians, with π\pi⁠: write π6\frac{\pi}{6}⁠, not 30∘30^\circ or 0.52.
  • Domain question, like cos⁡−1(3x−2)\cos^{-1}(3x - 2)⁠? The inside must lie in [−1,1][-1, 1]⁠. Solve −1≤3x−2≤1-1 \le 3x - 2 \le 1 to get 13≤x≤1\frac13 \le x \le 1⁠.
  • In an MCQ that asks for a principal value, first cross out every option outside the range.
  • Graphs: the graph of y=sin⁡−1xy = \sin^{-1}x is the graph of y=sin⁡xy = \sin x for x∈[−π2,π2]x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, mirrored in the line y=xy = x⁠. Domain and range swap.
Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

  • Writing cos⁡−1(−12)=−π3\cos^{-1}\left(-\frac12\right) = -\frac{\pi}{3} → cos⁡−1\cos^{-1} is never negative; the answer is 2π3\frac{2\pi}{3}⁠.
  • Writing sin⁡−1(sin⁡2π3)=2π3\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{2\pi}{3} → 2π3\frac{2\pi}{3} is outside the range; move the angle first and get π3\frac{\pi}{3}⁠.
  • Reading sin⁡−1x\sin^{-1}x as 1sin⁡x\frac{1}{\sin x} → sin⁡−1x\sin^{-1}x is an angle; 1sin⁡x\frac{1}{\sin x} is (sin⁡x)−1(\sin x)^{-1}⁠, a different thing.
  • Square brackets for the range of tan⁡−1\tan^{-1} → use round ones, (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, because tan⁡π2\tan\frac{\pi}{2} does not exist.
  • Using sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for any angle → it works only inside the range; check that first, on its own line.
Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

  • A CBSE sample-paper question (2025-26, not an NCERT exercise) split its 2 marks like this: half a mark for the line tan⁡(−π4+π3)\tan\left(-\frac{\pi}{4} + \frac{\pi}{3}\right)⁠, 1 mark for the tan⁡(A−B)\tan(A - B) formula and half a mark for the value.
  • Another CBSE sample-paper question (not an NCERT exercise) asked for the domain of cos⁡−1(3x−2)\cos^{-1}(3x - 2)⁠. Each inequality line earned half a mark, and so did the final interval.
  • In a "prove that" question, end with the statement itself and "Hence proved". The conclusion line carries marks.
  • Give angles exactly, as principal values in terms of π\pi⁠. There are no calculators, so no decimals.
  • An MCQ earns 1 mark for the right option only; the working earns nothing.
  • A slip is penalised only once, so keep going. Any correct method gets full marks.
Practice

Practise these first

Easiest first. Together they cover every type above.

  1. Exercise 2.1 Q3 a positive number in cosec⁡−1\operatorname{cosec}^{-1}
  2. Exercise 2.1 Q1 a negative number in sin⁡−1\sin^{-1}⁠: add a minus sign
  3. Exercise 2.1 Q9 a negative number in cos⁡−1\cos^{-1}⁠: take it away from π\pi
  4. Exercise 2.1 Q13 (MCQ) the range of sin⁡−1\sin^{-1}
  5. Exercise 2.1 Q11 adding three principal values
  6. Exercise 2.2 Q15 (MCQ) a difference of principal values, with cot⁡−1\cot^{-1} of a negative number
  7. Exercise 2.2 Q11 tan⁡−1(tan⁡…)\tan^{-1}(\tan\ldots) with the angle outside the range
  8. Exercise 2.2 Q13 (MCQ) cos⁡−1(cos⁡…)\cos^{-1}(\cos\ldots) with the angle outside the range
  9. Miscellaneous Q1 cos⁡−1(cos⁡…)\cos^{-1}(\cos\ldots) with an angle bigger than 2π2\pi
  10. Exercise 2.2 Q14 (MCQ) sin of an expression with an inverse: inside out
  11. Miscellaneous Q3 a "prove that" with the triangle trick
Next

You can do this

Six ranges and seven moves, one at a time. Work through the practice list, check every answer against its range, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

18 CBSE board and sample-paper questions on Inverse Trigonometric Functions, each with a full-marks exam answer. Practise them →