MathsCBSE Class 12 · Inverse Trigonometric Functions

Principal Values of Inverse Trigonometric Functions (Class 12)

Updated 15 October 2026 · Fact-checked

The principal value of an inverse trigonometric function is the one angle in its principal branch whose trigonometric value equals the input. Find the reference angle from the positive input, then place it in the branch: sin⁻¹, tan⁻¹ and cosec⁻¹ give a negative angle for a negative input; cos⁻¹, cot⁻¹ and sec⁻¹ give π minus the reference angle.

For a negative input, find the reference angle α, then the sin family gives −α while the cos family gives π − α.

Illustration drawn for QuestionBuddy.

Understand Principal Values of Inverse Trigonometric Functions

Many angles have the same sine: $sin \frac{π}{6} = sin \frac{5 π}{6} = sin (- \frac{7 π}{6}) = \frac{1}{2}$ . The principal value of $sin^{- 1} \frac{1}{2}$ is the one that lies in the principal branch $[- \frac{π}{2}, \frac{π}{2}]$ , namely $\frac{π}{6}$ . Every inverse trigonometric function returns exactly one value, always from its own branch:

$sin^{- 1}$ : $[- \frac{π}{2}, \frac{π}{2}]$ ; $tan^{- 1}$ : $(- \frac{π}{2}, \frac{π}{2})$ ; $cosec^{- 1}$ : $[- \frac{π}{2}, \frac{π}{2}] - {0}$ .
$cos^{- 1}$ : $[0, π]$ ; $cot^{- 1}$ : $(0, π)$ ; $sec^{- 1}$ : $[0, π] - {\frac{π}{2}}$ .

Positive inputs are easy: every branch contains the first-quadrant angles, so the answer is the familiar acute angle. $cos^{- 1} \frac{3}{2} = \frac{π}{6}$ , $tan^{- 1} 3 = \frac{π}{3}$ , $sec^{- 1} 2 = \frac{π}{3}$ , $cot^{- 1} 1 = \frac{π}{4}$ .

Negative inputs need the branch. First find the reference angle $α$ , the acute angle for the positive input. Then:

For the sin family ( $sin^{- 1}$ , $tan^{- 1}$ , $cosec^{- 1}$ ), the branch is centred on 0, so the answer is in the fourth quadrant: $- α$ . Example: $sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ , $tan^{- 1} (- 1) = - \frac{π}{4}$ , $cosec^{- 1} (- 2) = - \frac{π}{6}$ .
For the cos family ( $cos^{- 1}$ , $cot^{- 1}$ , $sec^{- 1}$ ), the branch runs from 0 to $π$ , so the answer is in the second quadrant: $π - α$ . Example: $cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ , $cot^{- 1} (- 3) = π - \frac{π}{6} = \frac{5 π}{6}$ , $sec^{- 1} (- 2) = \frac{2 π}{3}$ .

You can always check an answer two ways: the angle must lie in the branch, and the trigonometric function of the angle must equal the input. For instance, $cos \frac{2 π}{3} = - \frac{1}{2}$ and $\frac{2 π}{3} \in [0, π]$ , so $cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ is right, while $- \frac{π}{3}$ is wrong twice over: it lies outside $[0, π]$ , and $cos (- \frac{π}{3}) = \frac{1}{2}$ , not $- \frac{1}{2}$ .

Board questions combine several principal values in one expression, such as $cot^{- 1} (- 3) + cosec^{- 1} (- 2)$ , or put a principal value inside a trigonometric function, such as $cos (\frac{π}{6} + cos^{- 1} (- \frac{3}{2}))$ . In both, find each principal value separately on its own line, then combine. Do not use identities such as $cos^{- 1} (- x) = π - cos^{- 1} x$ as quoted properties: properties of inverse trigonometric functions are not in the 2026-27 syllabus, so reason from the branch each time.

The standard values you need come from the table of $0, \frac{π}{6}, \frac{π}{4}, \frac{π}{3}, \frac{π}{2}$ . NCERT covers principal values in Section 2.2 of Chapter 2.

Principal Values of Inverse Trigonometric Functions: formulas to remember

Principal value: $sin^{- 1} x = θ ⟺ sin θ = x and θ \in [- \frac{π}{2}, \frac{π}{2}]$; Same pattern for each function, with its own branch.
Negative input, sin family: $sin^{- 1} (- a), tan^{- 1} (- a), cosec^{- 1} (- a) \to - α$; $α$ is the reference angle for the positive input $a$ .
Negative input, cos family: $cos^{- 1} (- a), cot^{- 1} (- a), sec^{- 1} (- a) \to π - α$; The answer lies in the second quadrant.
Key sine and cosine values: $sin \frac{π}{6} = cos \frac{π}{3} = \frac{1}{2}, sin \frac{π}{4} = cos \frac{π}{4} = \frac{1}{2}, sin \frac{π}{3} = cos \frac{π}{6} = \frac{3}{2}$; Read reference angles from these.
Key tangent values: $tan \frac{π}{6} = \frac{1}{3}, tan \frac{π}{4} = 1, tan \frac{π}{3} = 3$; $cot$ values are the reciprocals.
Values at the ends: $sin^{- 1} 1 = \frac{π}{2}, sin^{- 1} (- 1) = - \frac{π}{2}, cos^{- 1} 1 = 0, cos^{- 1} (- 1) = π$; Also $sec^{- 1} 1 = 0$ , $sec^{- 1} (- 1) = π$ .

How to solve Principal Values of Inverse Trigonometric Functions questions

Use this for any 'find the principal value' or 'evaluate' question.

1Split the expression into single inverse terms and handle each on its own line.
2For each term, ignore the sign of the input and find the acute reference angle $α$ from the standard table. For $sec^{- 1} a$ and $cosec^{- 1} a$ , use $cos α = \frac{1}{a}$ or $sin α = \frac{1}{a}$ .
3If the input is positive, the principal value is $α$ .
4If the input is negative, use $- α$ for $sin^{- 1}$ , $tan^{- 1}$ , $cosec^{- 1}$ , and $π - α$ for $cos^{- 1}$ , $cot^{- 1}$ , $sec^{- 1}$ .
5Check each value: it lies in the right branch, and the trigonometric function of it gives the input.
6Combine the values with the given signs and coefficients; use a common denominator such as 12 for sixths and quarters.
7If the principal value sits inside an outer trigonometric function, substitute it and evaluate that function last.

Principal Values of Inverse Trigonometric Functions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Principal branch of inverse sine

Suppose a phone's tilt app reads a number $r$ between $- 1$ and $1$ from its sensor and displays the tilt angle $θ = sin^{- 1} r$ , positive when the top of the phone is raised and negative when it is lowered.

Find the displayed angle for readings $r = \frac{1}{2}$ , $r = - \frac{3}{2}$ and $r = - 1$ . Explain why the app never displays $\frac{7 π}{6}$ , even though $sin \frac{7 π}{6} = - \frac{1}{2}$ .

Try it first, then show the solutionSolution

$r = \frac{1}{2}$ : $sin^{- 1} \frac{1}{2} = \frac{π}{6}$ (raised by $30°$ ).
$r = - \frac{3}{2}$ : reference angle $\frac{π}{3}$ ; sin family with negative input gives $- \frac{π}{3}$ (lowered by $60°$ ).
$r = - 1$ : $sin^{- 1} (- 1) = - \frac{π}{2}$ (pointing straight down).
$sin^{- 1}$ always returns a value in $[- \frac{π}{2}, \frac{π}{2}]$ , and $\frac{7 π}{6}$ is outside it. For $r = - \frac{1}{2}$ the app shows $- \frac{π}{6}$ .

Answer: $\frac{π}{6}$ , $- \frac{π}{3}$ and $- \frac{π}{2}$ . The display is always in $[- \frac{π}{2}, \frac{π}{2}]$ , so $\frac{7 π}{6}$ never appears.

Memory hook: A tilt meter swings between straight down and straight up: that is the sine branch.

Principal branch of inverse cosine

Imagine a navigation app that measures the angle between a boat's heading and due north as $θ = cos^{- 1} c$ , where $c$ is a number between $- 1$ and $1$ computed from the heading. The angle is always between $0$ (heading north) and $π$ (heading south).

Find $θ$ for $c = \frac{1}{2}$ , $c = - \frac{1}{2}$ and $c = - 1$ . Why can the answer never be negative?

Try it first, then show the solutionSolution

$c = \frac{1}{2}$ : $cos^{- 1} \frac{1}{2} = \frac{π}{3}$ .
$c = - \frac{1}{2}$ : reference angle $\frac{π}{4}$ ; cos family with negative input gives $π - \frac{π}{4} = \frac{3 π}{4}$ .
$c = - 1$ : $cos^{- 1} (- 1) = π$ (heading due south).
The branch of $cos^{- 1}$ is $[0, π]$ , so every value is between 0 and $π$ and never negative.

Answer: $\frac{π}{3}$ , $\frac{3 π}{4}$ and $π$ ; values of $cos^{- 1}$ always lie in $[0, π]$ .

Memory hook: Angle from north runs from 0 to $π$ , like $cos^{- 1}$ : a negative input just means the boat points more south than north.

Tricks to remember Principal Values of Inverse Trigonometric Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Negative sin family goes below, cos family goes behind
For a negative input, the sin family drops below the x-axis ( $- α$ ); the cos family swings behind the y-axis ( $π - α$ ).
Use it when: Every negative-input principal value.
The cos family is never negative
If your answer for $cos^{- 1}$ , $cot^{- 1}$ or $sec^{- 1}$ is negative, it is wrong.
Use it when: A last-second check on MCQs.
Reciprocal first for sec and cosec
$sec^{- 1} a$ is the angle whose cosine is $\frac{1}{a}$ ; $cosec^{- 1} a$ is the angle whose sine is $\frac{1}{a}$ . Flip, then use the cos or sin table, then apply the branch.
Use it when: Terms like $sec^{- 1} (\frac{2}{3})$ and $cosec^{- 1} (- 2)$ .
Common denominator 12
Sixths, quarters and thirds all go to twelfths: $\frac{π}{6} = \frac{2 π}{12}$ , $\frac{π}{4} = \frac{3 π}{12}$ , $\frac{π}{3} = \frac{4 π}{12}$ .
Use it when: Adding several principal values without slips.

Principal Values of Inverse Trigonometric Functions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two negative inputs from different families · 2 marks · Very short answerBoard level

Find the value of $cot^{- 1} (- 3) + cosec^{- 1} (- 2)$ .

Try it first, then show the solutionSolution

$cot \frac{π}{6} = 3$ , so the reference angle is $\frac{π}{6}$ . $cot^{- 1}$ is in the cos family, so $cot^{- 1} (- 3) = π - \frac{π}{6} = \frac{5 π}{6}$ . (1 mark)
$cosec \frac{π}{6} = 2$ ; $cosec^{- 1}$ is in the sin family, so $cosec^{- 1} (- 2) = - \frac{π}{6}$ . (½ mark)
Sum $= \frac{5 π}{6} - \frac{π}{6} = \frac{4 π}{6} = \frac{2 π}{3}$ . (½ mark)

Answer: $\frac{2 π}{3}$

Example 2: A principal value inside a cosine · 2 marks · Very short answerBoard level

Evaluate $cos (\frac{π}{6} + cos^{- 1} (- \frac{3}{2}))$ .

Try it first, then show the solutionSolution

Reference angle for $\frac{3}{2}$ under cosine is $\frac{π}{6}$ , so $cos^{- 1} (- \frac{3}{2}) = π - \frac{π}{6} = \frac{5 π}{6}$ . (1 mark)
The expression becomes $cos (\frac{π}{6} + \frac{5 π}{6}) = cos π = - 1$ . (1 mark)

Answer: $- 1$

Example 3: Three branches in one expression · 3 marks · Short answerHarder board level

Find the value of $sec^{- 1} (- 2) - cosec^{- 1} (2) + cot^{- 1} (- 1)$ .

Try it first, then show the solutionSolution

$sec^{- 1} (- 2)$ : $cos α = \frac{1}{2}$ gives $α = \frac{π}{4}$ ; cos family, negative input: $π - \frac{π}{4} = \frac{3 π}{4}$ . (1 mark)
$cosec^{- 1} (2) = \frac{π}{4}$ , since $sin \frac{π}{4} = \frac{1}{2}$ and the input is positive. (½ mark)
$cot^{- 1} (- 1)$ : reference angle $\frac{π}{4}$ ; cos family: $\frac{3 π}{4}$ . (1 mark)
Value $= \frac{3 π}{4} - \frac{π}{4} + \frac{3 π}{4} = \frac{5 π}{4}$ . (½ mark)

Answer: $\frac{5 π}{4}$

Example 4: Mixed coefficients · 3 marks · Short answerHarder board level

Evaluate $3 sin^{- 1} (\frac{1}{2}) - 2 cos^{- 1} (- \frac{1}{2}) + tan^{- 1} (- \frac{1}{3})$ .

Try it first, then show the solutionSolution

$sin^{- 1} \frac{1}{2} = \frac{π}{4}$ . (½ mark)
$cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{4} = \frac{3 π}{4}$ . (1 mark)
$tan^{- 1} (- \frac{1}{3}) = - \frac{π}{6}$ . (½ mark)
Value $= \frac{3 π}{4} - \frac{3 π}{2} - \frac{π}{6} = \frac{9 π - 18 π - 2 π}{12} = - \frac{11 π}{12}$ . (1 mark)

Answer: $- \frac{11 π}{12}$

Example 5: An equation solved with branches, not identities · 3 marks · Short answerHOTS / competency

Find all real $x$ such that $cos^{- 1} x = 2 sin^{- 1} x$ .

Try it first, then show the solutionSolution

Let $θ = sin^{- 1} x$ , so $x = sin θ$ with $θ \in [- \frac{π}{2}, \frac{π}{2}]$ . Then $cos^{- 1} x = 2 θ$ must lie in $[0, π]$ , so $θ \in [0, \frac{π}{2}]$ . (1 mark)
$cos^{- 1} x = 2 θ$ means $x = cos 2 θ = 1 - 2 sin^{2} θ$ . With $x = sin θ = s$ : $s = 1 - 2 s^{2}$ , so $2 s^{2} + s - 1 = 0$ , giving $(2 s - 1) (s + 1) = 0$ . (1 mark)
$s = - 1$ would need $θ = - \frac{π}{2}$ , outside $[0, \frac{π}{2}]$ ; so $s = \frac{1}{2}$ and $x = \frac{1}{2}$ . Check: $cos^{- 1} \frac{1}{2} = \frac{π}{3}$ and $2 sin^{- 1} \frac{1}{2} = \frac{π}{3}$ . (1 mark)

Answer: $x = \frac{1}{2}$

Example 6: Case study: angles of elevation from two points · 4 marks · Case-basedHOTS / competency

Suppose two friends A and B stand on level ground on the same side of a 30 m tall tower, in line with its foot. From A the angle of elevation of the top is $tan^{- 1} (3)$ , and from B it is $cot^{- 1} (3)$ .
(i) Find both angles as principal values. (1 mark)
(ii) Who is nearer the tower? (1 mark)
(iii) Find the distance AB. (2 marks)

Try it first, then show the solutionSolution

(i) $tan^{- 1} 3 = \frac{π}{3}$ and $cot^{- 1} 3 = \frac{π}{6}$ . (1 mark)
(ii) The larger angle of elevation belongs to the nearer observer, so A is nearer. (1 mark)
(iii) Distance of A from the foot $= \frac{30}{t a n \frac{π}{3}} = \frac{30}{3} = 103$ m; of B $= \frac{30}{t a n \frac{π}{6}} = 303$ m. (1 mark)
$A B = 303 - 103 = 203$ m $\approx 34.6$ m. (1 mark)

Answer: (i) $\frac{π}{3}$ and $\frac{π}{6}$ . (ii) A. (iii) $203$ m.

Where students lose marks in Principal Values of Inverse Trigonometric Functions

The slips examiners see most often, and what to write instead.

Mistake: Giving a negative answer for cos inverse
Students copy the sin pattern $- α$ to the cos family.
Wrong working
$cos^{- 1} (- \frac{1}{2}) = - \frac{π}{3}$ .
Right working
$cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ .
Do this: Write the branch next to each term before evaluating: ' $cos^{- 1}$ : $[0, π]$ '.
Usually costs: The value and every step that uses it.
Mistake: Answering with an angle outside the branch
Any angle with the right trigonometric value is accepted mentally, such as $\frac{7 π}{6}$ for $sin^{- 1} (- \frac{1}{2})$ .
Wrong working
$sin^{- 1} (- \frac{1}{2}) = \frac{7 π}{6}$ .
Right working
$sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ , the only such angle in $[- \frac{π}{2}, \frac{π}{2}]$ .
Do this: After finding an angle, confirm it lies in the branch.
Usually costs: 1 mark, or the MCQ.
Mistake: Not flipping the input for sec and cosec inverse
$sec^{- 1} 2$ is not the angle whose cosine is 2.
Wrong working
$sec^{- 1} (\frac{2}{3})$ : $cos θ = \frac{2}{3}$ , impossible.
Right working
$sec θ = \frac{2}{3} \Rightarrow cos θ = \frac{3}{2} \Rightarrow θ = \frac{π}{6}$ .
Do this: Always convert sec and cosec to cos and sin first.
Usually costs: The term's value.
Mistake: Quoting dropped properties as reasons
Properties like $cos^{- 1} (- x) = π - cos^{- 1} x$ are outside the 2026-27 syllabus; examiners expect branch reasoning.
Wrong working
$cot^{- 1} (- 3) = - cot^{- 1} 3 = - \frac{π}{6}$ .
Right working
$cot^{- 1} (- 3) = \frac{5 π}{6}$ , because $cot \frac{5 π}{6} = - 3$ and $\frac{5 π}{6} \in (0, π)$ .
Do this: Justify each value with 'value of the function' and 'lies in the branch'.
Usually costs: The value (here the sign is also wrong).

Exam tips

Principal-value MCQs and 2-mark evaluations appear in nearly every recent paper (2023 to 2026 board papers and the sample papers); write each principal value on its own line.
Keep exact forms such as $\frac{2 π}{3}$ ; do not convert to degrees unless asked.
For an expression inside $sin$ or $cos$ , evaluate the inverse part fully first, then the outer function.
Before submitting an MCQ answer, rule out options outside the branch: this often leaves only one choice.
If a 1-mark question gives a non-standard input like $sin^{- 1} (0.3)$ , it is testing the range or domain, not a numerical value.

Practice questions on Principal Values of Inverse Trigonometric Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The principal value of $cos^{- 1} (- \frac{1}{2})$ is:
(a) $- \frac{π}{3}$
(b) $\frac{π}{3}$
(c) $\frac{2 π}{3}$
(d) $\frac{4 π}{3}$
Hint
Cos family, negative input: $π - α$ .
Answer
(c) $\frac{2 π}{3}$ .
Q2 · 1 mark · MCQBoard level
The principal value of $cosec^{- 1} (- 2)$ is:
(a) $- \frac{π}{4}$
(b) $\frac{3 π}{4}$
(c) $\frac{5 π}{4}$
(d) $\frac{π}{4}$
Hint
Sin family: a negative input gives a negative angle.
Answer
(a) $- \frac{π}{4}$ , since $sin (- \frac{π}{4}) = - \frac{1}{2}$ and $- \frac{π}{4}$ is in the branch.
Q3 · 2 marks · Very short answerBoard level
Find the value of $tan^{- 1} (- 3) + cot^{- 1} (- 1)$ .
Hint
$tan^{- 1}$ is in the sin family, $cot^{- 1}$ in the cos family.
Answer
$tan^{- 1} (- 3) = - \frac{π}{3}$ and $cot^{- 1} (- 1) = \frac{3 π}{4}$ . Sum $= \frac{- 4 π + 9 π}{12} = \frac{5 π}{12}$ .
Q4 · 2 marks · Very short answerHarder board level
Evaluate $sin (\frac{π}{2} - sin^{- 1} (- \frac{3}{2}))$ .
Hint
Find $sin^{- 1} (- \frac{3}{2})$ first; watch the double negative.
Answer
$sin^{- 1} (- \frac{3}{2}) = - \frac{π}{3}$ , so the expression is $sin (\frac{π}{2} + \frac{π}{3}) = sin \frac{5 π}{6} = \frac{1}{2}$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The principal value of $sin^{- 1} (- \frac{1}{2})$ is $- \frac{π}{6}$ .
Reason (R): $sin (- \frac{π}{6}) = - \frac{1}{2}$ and $- \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}]$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
A principal value needs two things: the right sine value and the right branch.
Answer
(a). R states exactly the two conditions that define the principal value, so it explains A. (Angles like $\frac{7 π}{6}$ also have sine $- \frac{1}{2}$ but are outside the branch.)
Q6 · 3 marks · Short answerHarder board level
Find the value of $cot^{- 1} (\frac{1}{3}) - cosec^{- 1} (- \frac{2}{3}) + sec^{- 1} (\frac{2}{3})$ .
Hint
Flip the sec and cosec inputs to get $\frac{3}{2}$ .
Answer
$cot^{- 1} \frac{1}{3} = \frac{π}{3}$ ; $cosec^{- 1} (- \frac{2}{3}) = - \frac{π}{3}$ (sine $- \frac{3}{2}$ , sin family); $sec^{- 1} \frac{2}{3} = \frac{π}{6}$ (cosine $\frac{3}{2}$ ). Value $= \frac{π}{3} + \frac{π}{3} + \frac{π}{6} = \frac{5 π}{6}$ .
Q7 · 3 marks · Short answerHOTS / competency
Find all real $x$ for which $sin^{- 1} x = 2 cos^{- 1} x$ .
Hint
Let $φ = cos^{- 1} x \in [0, π]$ . Since $sin^{- 1} x = 2 φ$ must be at most $\frac{π}{2}$ , $φ \leq \frac{π}{4}$ .
Answer
With $x = cos φ$ and $φ \in [0, \frac{π}{4}]$ , the equation means $x = sin 2 φ = 2 sin φ cos φ$ . So $cos φ (1 - 2 sin φ) = 0$ . $cos φ = 0$ needs $φ = \frac{π}{2}$ , not allowed; so $sin φ = \frac{1}{2}$ , $φ = \frac{π}{6}$ , $x = cos \frac{π}{6} = \frac{3}{2}$ . Check: $sin^{- 1} \frac{3}{2} = \frac{π}{3} = 2 \cdot \frac{π}{6}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Principal value of $tan^{- 1} (- 1)$ ?Show answer
$- \frac{π}{4}$ .
2.Principal value of $sec^{- 1} (- 2)$ ?Show answer
$\frac{2 π}{3}$ (cosine $- \frac{1}{2}$ in $[0, π]$ ).
3.Principal value of $cot^{- 1} (- 3)$ ?Show answer
$\frac{5 π}{6}$ .
4.Principal value of $sin^{- 1} (- 1)$ ?Show answer
$- \frac{π}{2}$ .
5.Which three inverse functions never give a negative principal value?Show answer
$cos^{- 1}$ , $cot^{- 1}$ and $sec^{- 1}$ .
6.What is $cos^{- 1} (- \frac{1}{2})$ ?Show answer
$\frac{3 π}{4}$ .

Principal Values of Inverse Trigonometric Functions: frequently asked questions

What is the principal value of an inverse trigonometric function?

It is the single value that the function returns: the angle in its principal value branch whose trigonometric value equals the given number. For example, the principal value of sin inverse of 1/2 is π/6, because π/6 is in [−π/2, π/2] and sin(π/6) = 1/2.

How do I find the principal value for a negative input?

Find the acute reference angle α for the positive input. For sin inverse, tan inverse and cosec inverse, the answer is −α. For cos inverse, cot inverse and sec inverse, the answer is π − α.

Can I use sin inverse of (−x) = −sin inverse of x?

Properties of inverse trigonometric functions are not in the 2026-27 syllabus. The result is still true, but justify each value from the branch: state the angle, check its trigonometric value, and check that it lies in the branch.

Which NCERT section has principal value questions?

Section 2.2 of Chapter 2, Inverse Trigonometric Functions, in NCERT Mathematics Part I for Class 12, together with the exercise that follows it.

Are answers in degrees accepted?

Inverse trigonometric values are normally written in radians as multiples of π, such as 2π/3, and MCQ options are given that way. Use radians unless the question asks for degrees.