Is it correct to say: Writing cos⁻¹(−1/2) = −π/3?

The range of cos⁻¹ is [0, π], so a negative angle is impossible. cos⁻¹(−1/2) = π − π/3 = 2π/3.

Is it correct to say: Writing sin⁻¹(sin 2π/3) = 2π/3?

2π/3 lies outside [−π/2, π/2]. Since sin(2π/3) = sin(π/3), the value is π/3.

Is it correct to say: Using cos⁻¹(−x) = −cos⁻¹x?

That rule holds for sin⁻¹, tan⁻¹ and cosec⁻¹. For cos⁻¹, sec⁻¹ and cot⁻¹ the correct rule is π minus the angle, e.g. cos⁻¹(−x) = π − cos⁻¹x.

Is it correct to say: Writing cot⁻¹(−√3) = −π/6?

The range of cot⁻¹ is (0, π). cot(5π/6) = −√3, so cot⁻¹(−√3) = 5π/6.

Is it correct to say: Reading sin⁻¹x as 1/sin x?

1/sin x is (sin x)⁻¹ = cosec x. sin⁻¹x is the angle whose sine is x.

Is it correct to say: Evaluating sec⁻¹(1/2) or cosec⁻¹(0.5)?

sec⁻¹ and cosec⁻¹ are defined only for |x| ≥ 1. sec⁻¹(1/2) does not exist.

📄 Formula sheet (PDF)

Inverse Trigonometric Functions

Domains, ranges and principal value branches of the six inverse trigonometric functions, principal values, and evaluating expressions such as sin⁻¹(sin x) — NCERT Class 12 Maths Ch 2

📐 18 formulas✏️ 4 examples🎯 6 practice⚖️ Unit I: 8 marks (with Ch 1)🏫 CBSE📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Learn the table of six principal value branches first. Almost every question in this chapter depends on it.
→A principal value must lie inside the branch. cos⁻¹(−1/2) is 2π/3, not −π/3 and not 4π/3.
→sin⁻¹(sin x) = x only when x lies in [−π/2, π/2]. Otherwise, first rewrite sin x as the sine of an angle inside the branch.
→For a domain question such as sin⁻¹(2x − 3), solve −1 ≤ 2x − 3 ≤ 1. With two inverse functions, intersect the two conditions.
→For 'simplest form' questions, substitute x = a sin θ, a tan θ or a sec θ, simplify inside the bracket, then check that the angle lies in the principal branch.
→sin⁻¹x means the inverse sine (arcsine). It is not 1/sin x.

📐 Formulas(18)

Principal Branch of sin⁻¹★ Board fav

sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]

Principal Branch of cos⁻¹★ Board fav

cos^{- 1} : [- 1, 1] \to [0, π]

Principal Branch of tan⁻¹★ Board fav

tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2})

Principal Branch of cot⁻¹

cot^{- 1} : R \to (0, π)

Principal Branch of sec⁻¹

sec^{- 1} : R - (- 1, 1) \to [0, π] - {\frac{π}{2}}

Principal Branch of cosec⁻¹

cosec^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}] - {0}

Meaning of an Inverse Trig Value

y = sin^{- 1} x \Leftrightarrow sin y = x, y \in [- \frac{π}{2}, \frac{π}{2}]

Notation Warning

sin^{- 1} x \neq = (sin x)^{- 1} = \frac{1}{s i n x}

Function of its Inverse

sin (sin^{- 1} x) = x, cos (cos^{- 1} x) = x (- 1 \leq x \leq 1); tan (tan^{- 1} x) = x (x \in R)

Inverse of the Function (sin, tan)★ Board fav

sin^{- 1} (sin x) = x (- \frac{π}{2} \leq x \leq \frac{π}{2}), tan^{- 1} (tan x) = x (- \frac{π}{2} < x < \frac{π}{2})

Inverse of the Function (cos, cot)

cos^{- 1} (cos x) = x (0 \leq x \leq π), cot^{- 1} (cot x) = x (0 < x < π)

sin⁻¹(sin x) Outside the Branch★ Board fav

sin^{- 1} (sin x) = π - x, \frac{π}{2} \leq x \leq \frac{3 π}{2}

cos⁻¹(cos x) Outside the Branch

cos^{- 1} (cos x) = 2 π - x, π \leq x \leq 2 π

tan⁻¹(tan x) Outside the Branch

tan^{- 1} (tan x) = x - π, \frac{π}{2} < x < \frac{3 π}{2}

Negative Arguments: sin⁻¹, tan⁻¹, cosec⁻¹

sin^{- 1} (- x) = - sin^{- 1} x, tan^{- 1} (- x) = - tan^{- 1} x, cosec^{- 1} (- x) = - cosec^{- 1} x

Negative Arguments: cos⁻¹, cot⁻¹, sec⁻¹★ Board fav

cos^{- 1} (- x) = π - cos^{- 1} x, cot^{- 1} (- x) = π - cot^{- 1} x, sec^{- 1} (- x) = π - sec^{- 1} x

Domain Condition for sin⁻¹ and cos⁻¹

sin^{- 1} (g (x)), cos^{- 1} (g (x)) defined \Leftrightarrow - 1 \leq g (x) \leq 1

Substitutions for Simplest Form

a^{2} - x^{2} : x = a sin θ; a^{2} + x^{2} : x = a tan θ; x^{2} - a^{2} : x = a sec θ

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the principal value of cos⁻¹(−√3/2).

Let y = cos⁻¹(−√3/2). Then cos y = −√3/2 with y in [0, π].

cos y = - \frac{3}{2}, y \in [0, π]

2Solved Exampleboard4 steps

Find the value of cot⁻¹(−1) + cosec⁻¹(−√2) + sec⁻¹(2).

cot⁻¹ has range (0, π). cot(π/4) = 1, so for −1 take π − π/4.

cot^{- 1} (- 1) = π - \frac{π}{4} = \frac{3 π}{4}

3Solved Exampleboard4 steps

Find the value of sin⁻¹(sin 4π/5) + cos⁻¹(cos 5π/3).

4π/5 is NOT in [−π/2, π/2]. Use sin x = sin(π − x) to bring it into the branch.

sin \frac{4 π}{5} = sin (π - \frac{4 π}{5}) = sin \frac{π}{5}, \frac{π}{5} \in [- \frac{π}{2}, \frac{π}{2}]

4Solved ExampleHOTS4 steps

Write tan⁻¹ √((1 + cos x)/(1 − cos x)), 0 < x < π, in the simplest form.

Use the half-angle identities 1 + cos x = 2cos²(x/2) and 1 − cos x = 2sin²(x/2).

\frac{1 + c o s x}{1 - c o s x} = \frac{2 c o s ^{2} ( x /2 )}{2 s i n ^{2} ( x /2 )} = cot^{2} \frac{x}{2}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing cos⁻¹(−1/2) = −π/3
✓The range of cos⁻¹ is [0, π], so a negative angle is impossible. cos⁻¹(−1/2) = π − π/3 = 2π/3.
2
Writing sin⁻¹(sin 2π/3) = 2π/3
✓2π/3 lies outside [−π/2, π/2]. Since sin(2π/3) = sin(π/3), the value is π/3.
3
Using cos⁻¹(−x) = −cos⁻¹x
✓That rule holds for sin⁻¹, tan⁻¹ and cosec⁻¹. For cos⁻¹, sec⁻¹ and cot⁻¹ the correct rule is π minus the angle, e.g. cos⁻¹(−x) = π − cos⁻¹x.
4
Writing cot⁻¹(−√3) = −π/6
✓The range of cot⁻¹ is (0, π). cot(5π/6) = −√3, so cot⁻¹(−√3) = 5π/6.
5
Reading sin⁻¹x as 1/sin x
✓1/sin x is (sin x)⁻¹ = cosec x. sin⁻¹x is the angle whose sine is x.
6
Evaluating sec⁻¹(1/2) or cosec⁻¹(0.5)
✓sec⁻¹ and cosec⁻¹ are defined only for |x| ≥ 1. sec⁻¹(1/2) does not exist.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the principal value of tan⁻¹(−1/√3).
Q2
Find the principal value of sec⁻¹(−√2).
Q3
Find the domain of f(x) = sin⁻¹(2x − 3) + cos⁻¹(x/2).
Q4
Find the value of cos⁻¹(cos 7π/4).
Q5
Find the value of tan⁻¹(tan 4π/3).
Q6
Find the value of cos(π/6 + cos⁻¹(−1/2)).

📝 Notes

Inverse Trigonometric Functions

Trigonometric functions repeat their values, so they are not one-one on R and have no inverse there. Chapter 2 restricts each function to one interval (for sec and cosec, an interval with one point removed), its principal value branch, on which it is one-one and onto. The values of the inverse function lie in that branch.

Read the range table first

sin⁻¹x: domain [−1, 1], range [−π/2, π/2]
cos⁻¹x: domain [−1, 1], range [0, π]
tan⁻¹x: domain R, range (−π/2, π/2)
cot⁻¹x: domain R, range (0, π)
sec⁻¹x: domain R − (−1, 1), range [0, π] − {π/2}
cosec⁻¹x: domain R − (−1, 1), range [−π/2, π/2] − {0}

The ranges of sin⁻¹, tan⁻¹ and cosec⁻¹ are symmetric about 0, so negative inputs give negative angles. The ranges of cos⁻¹, cot⁻¹ and sec⁻¹ lie in [0, π], so negative inputs give obtuse angles.

Values outside the branch

An expression like sin⁻¹(sin x) equals x only when x is already in the branch. If it is not, rewrite the inner function using an identity that does not change its value, such as sin x = sin(π − x), cos x = cos(2π − x) or tan x = tan(x − π), until the angle lies in the branch. Then read off the answer.

Simplest-form questions

Choose a substitution suggested by the expression (x = a sin θ, a tan θ, a sec θ), or use a half-angle identity for 1 ± cos x.
Simplify inside the inverse function until it is f(θ) for a single angle θ.
Check that θ lies in the principal branch for the given range of x, then write f⁻¹(f(θ)) = θ and go back to x.

The third step is where marks are lost. Always state the interval in which θ lies.

This page covers the principal-value branches and the identities f(f⁻¹(x)) = x and f⁻¹(f(x)) = x that the current NCERT textbook states, together with results that follow directly from them.