Inverse Trigonometric Functions: Class 12 Maths · Chapter 2

Mathematics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

65 min
48 practice questions
16 MCQs

What you will learn in Inverse Trigonometric Functions

Explain why each trigonometric function must be restricted before it has an inverse, and name the interval used for its principal value branch.
State the domain and principal range of each of the six inverse trigonometric functions and sketch its graph by reflecting a branch in the line y = x.
Find principal values of inverse trigonometric functions at standard arguments, including negative ones.
Use sin(sin⁻¹x) = x and sin⁻¹(sin x) = x, and their analogues, only on the intervals where they hold, and evaluate expressions such as sin⁻¹(sin a) for any angle a.
Simplify inverse trigonometric expressions with a suitable substitution, keeping track of the interval of the new angle.
Evaluate and prove sums and compositions of inverse trigonometric functions using right triangles and addition formulae, and solve simple equations involving them.

Original companion text aligned to NCERT Mathematics Class 12, Chapter 2, pp. 18–33. Not affiliated with or endorsed by NCERT.

2.1 Introduction

Tilt a phone and its screen can show the tilt in degrees. The accelerometer inside does not measure that angle. It measures what fraction of gravity acts along the phone's long edge, and that fraction is the sine of the tilt. To show degrees, the software must run sine backwards: from $\sin θ = 0.5$ it has to recover $θ$ .

Running a function backwards is the job of its inverse, and Chapter 1 told us when that job can be done: the function has to be one-one and onto. Sine fails the first test badly: $\sin θ = 0.5$ for $θ = \frac{π}{6}$ , $\frac{5 π}{6}$ , $\frac{13 π}{6}$ and infinitely many other angles (Fig. 2.1). It is a many-one function, and so are the other five trigonometric functions.

4/4On $- \frac{π}{2} \leq x \leq \frac{π}{2}$ there is exactly one answer, $x = \frac{π}{6}$ .

Fig. 2.1One output of sine comes from many inputs, until the domain is cut down.

Step 1 shows the graph of y = sin x from about minus 2 pi to 2 pi, a wave between minus 1 and 1. Step 2 adds a dashed horizontal line at height one half. Step 3 marks the four places where the line meets the wave, with the words many angles. Step 4 fades the wave except the piece from minus pi by 2 to pi by 2, shades that strip, and keeps a single crossing at x = pi by 6, labelled one angle.

The cure is to throw part of the domain away. If we agree to look only at angles from $- \frac{π}{2}$ to $\frac{π}{2}$ , exactly one of them has sine $0.5$ , and the backwards question has a single answer. This chapter makes such a choice for all six trigonometric functions, draws the inverse functions that result and works out the rules they obey.

These inverses are not just a patch for a technicality. Calculus produces them as the answers to integrals such as $\int \frac{d x}{1 + x^{2}}$ , and surveyors, physicists and engineers reach for them whenever an angle has to be recovered from measured lengths: the gradient of a road, the elevation of a dish antenna, the joint angles of a robot arm.

Quick check

1 of 2

Why does

\sin x

, taken on all real numbers, have no inverse?

Its graph is a curve, not a straight lineIt takes negative values for some anglesDifferent angles give the same sine, so it is not one-oneIt is undefined at

x = \frac{π}{2}

Show answer

Answer (C) Different angles give the same sine, so it is not one-one

An inverse must send each output back to one input, but

\sin \frac{π}{6} = \sin \frac{5 π}{6}

. Curved graphs and negative values are no obstacle (think of

x^{3}

), and sine is defined at

\frac{π}{2}

; it is tangent that is not.

2.2 Basic Concepts

Two facts from earlier classes do all the work in this section: where each trigonometric function lives, and what it takes to invert a function. The first is collected in Table 2.1.

Table 2.1The six trigonometric functions on their natural domains

Function	Domain	Range
$\sin x$	$𝐑$	$[- 1, 1]$
$\cos x$	$𝐑$	$[- 1, 1]$
$\tan x$	$𝐑 - {(2 n + 1) \frac{π}{2} : n \in 𝐙}$	$𝐑$
$\cot x$	$𝐑 - {n π : n \in 𝐙}$	$𝐑$
$\sec x$	$𝐑 - {(2 n + 1) \frac{π}{2} : n \in 𝐙}$	$𝐑 - (- 1, 1)$
$cosec x$	$𝐑 - {n π : n \in 𝐙}$	$𝐑 - (- 1, 1)$

Each of the six repeats itself, with period $2 π$ or $π$ , so none is one-one on its natural domain. Now the second fact.

Definition

Inverse function

The function $f^{- 1}$ that undoes a one-one and onto function $f$ : it sends each output $f (x)$ back to the input $x$ . Inputs and outputs trade places, so the domain of $f^{- 1}$ is the range of $f$ , and the other way round.

Example $f (x) = 2 x + 1$ on $𝐑$ has inverse $f^{- 1} (y) = \frac{y - 1}{2}$ .

Undo, then redo

\begin{array}{c} f^{- 1} (f (x)) = x \\ f (f^{- 1} (y)) = y \end{array}

These hold for every $x$ in $X$ and every $y$ in $Y$ . Undoing works in both directions, so $f$ is in turn the inverse of $f^{- 1}$ . That fixes the plan for every trigonometric function: pick an interval on which it is one-one and still takes every value of its range, then invert that piece.

Inverse sine

On $[- \frac{π}{2}, \frac{π}{2}]$ the sine function climbs steadily from $- 1$ to $1$ and passes every value in between exactly once. Restricted to this interval, $\sin : [- \frac{π}{2}, \frac{π}{2}] \to [- 1, 1]$ is one-one and onto, so it has an inverse. Such a cut-down interval is called a restricted domain.

It is not the only interval that works. Sine falls steadily from $1$ to $- 1$ on $[\frac{π}{2}, \frac{3 π}{2}]$ and climbs again on $[- \frac{5 π}{2}, - \frac{3 π}{2}]$ ; in fact any interval $[(2 k - 1) \frac{π}{2}, (2 k + 1) \frac{π}{2}]$ with $k$ an integer will do. An interval such as $[0, π]$ does not work, because the curve rises and then comes back down over it (Fig. 2.2).

2/4On $[- \frac{π}{2}, \frac{π}{2}]$ sine rises from $- 1$ to $1$ : one-one and onto $[- 1, 1]$ .

Fig. 2.2Which pieces of the sine curve can be inverted?NCERT Fig. 2.1 · p. 20

Step 1 shows the sine curve from minus 2 pi to 2 pi. Step 2 shades the strip from minus pi by 2 to pi by 2 and draws that piece thick and blue, rising from minus 1 to 1, tagged principal piece. Step 3 instead highlights the piece from pi by 2 to 3 pi by 2 in orange, falling from 1 to minus 1, tagged also one-one. Step 4 highlights the piece from 0 to pi with a dotted red line and a horizontal line at height one half that meets it twice, tagged two inputs: fails.

Each admissible interval gives a different inverse, called a branch of $\sin^{- 1}$ (read “sine inverse” or “arc sine”). Every branch has domain $[- 1, 1]$ ; the branches differ in their range. By agreement we use the branch whose range is $[- \frac{π}{2}, \frac{π}{2}]$ . It is the principal value branch of $\sin^{- 1}$ .

Definition

Principal value branch

Among the intervals on which a trigonometric function is one-one and onto, the one fixed by convention as the range of its inverse. For $\sin^{- 1}$ it is $[- \frac{π}{2}, \frac{π}{2}]$ .

ExampleThe branch of $\sin^{- 1}$ with range $[\frac{π}{2}, \frac{3 π}{2}]$ is a valid branch, but not the principal one.

Inverse sine

\begin{array}{c} \sin^{- 1} : [- 1, 1] \\ \to [- \frac{π}{2}, \frac{π}{2}] \end{array}

In words, $y = \sin^{- 1} x$ means that $\sin y = x$ and $- \frac{π}{2} \leq y \leq \frac{π}{2}$ . This is our first inverse trigonometric function, and the condition on $y$ is part of its definition. For example $\sin \frac{5 π}{6} = \frac{1}{2}$ , yet $\sin^{- 1} \frac{1}{2} = \frac{π}{6}$ and not $\frac{5 π}{6}$ , because only $\frac{π}{6}$ lies in the principal branch. Two facts follow at once. First, $\sin (\sin^{- 1} x) = x$ for every $x \in [- 1, 1]$ . Second, $\sin^{- 1} (\sin x) = x$ for every $x \in [- \frac{π}{2}, \frac{π}{2}]$ , and only there.

Drawing the graph. Suppose $f$ is invertible and $f (u) = v$ . Then $f^{- 1} (v) = u$ , so whenever $(u, v)$ is on the graph of $f$ , the point $(v, u)$ is on the graph of $f^{- 1}$ . Swapping the coordinates of every point is the same as reflecting the plane in the line $y = x$ . So the graph of $y = \sin^{- 1} x$ is the mirror image, in that line, of the principal piece of $y = \sin x$ (Fig. 2.3).

3/3Every point reflects the same way. The image is $y = \sin^{- 1} x$ , from $(- 1, - \frac{π}{2})$ to $(1, \frac{π}{2})$ .

Fig. 2.3Reflecting in

y = x

swaps the coordinates of every point.NCERT Fig. 2.1 (iii) · p. 20

Equal scales on both axes, from about minus 2 to 2. Step 1 shows the blue piece of y = sin x from minus pi by 2 to pi by 2, with the end point P at (pi by 2, 1). Step 2 adds the dashed mirror line y = x and the image point P prime at (1, pi by 2), joined to P by a dotted segment across the mirror. Step 3 adds the whole mirror image as a dashed red curve, y = sin inverse x, running from (minus 1, minus pi by 2) to (1, pi by 2), steep at both ends.

Read the new graph like any other. It exists only for $- 1 \leq x \leq 1$ , it rises from $- \frac{π}{2}$ to $\frac{π}{2}$ , and it is steepest at its ends, exactly where the sine curve was flattest. Any other branch of $\sin^{- 1}$ would be the mirror image of a different piece, sitting higher or lower on the $y$ -axis.

Inverse cosine

Cosine is an even function, so it cannot be one-one on any interval that is symmetric about $0$ . Instead we use $[0, π]$ , on which cosine falls steadily from $1$ to $- 1$ . Other admissible intervals are $[π, 2 π]$ , $[- π, 0]$ and, in general, $[k π, (k + 1) π]$ ; each gives a branch of $\cos^{- 1}$ with domain $[- 1, 1]$ .

Inverse cosine

\begin{array}{c} \cos^{- 1} : [- 1, 1] \\ \to [0, π] \end{array}

So $y = \cos^{- 1} x$ means that $\cos y = x$ and $0 \leq y \leq π$ . Reflecting the piece in $y = x$ gives a curve that falls from $(- 1, π)$ to $(1, 0)$ (Fig. 2.4). Notice that $\cos^{- 1}$ of a negative number is an obtuse angle, never a negative one: the whole range $[0, π]$ is non-negative.

3/3Go up from $x = - \frac{1}{2}$ to the curve and across: $\cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ .

Fig. 2.4The principal branch of

\cos^{- 1}

takes values from

0

π

.NCERT Fig. 2.2 · p. 21

Equal scales on both axes. Step 1 shows the whole cosine curve as a faint dotted line, with the piece from x = 0 to x = pi drawn thick and blue, falling from 1 to minus 1. Step 2 adds the dashed mirror line y = x and the red dashed image y = cos inverse x, which falls from (minus 1, pi) to (1, 0). Step 3 marks the point of the image above x = minus one half, with dotted guides to both axes; it sits at height 2 pi by 3.

Inverse cosecant and secant

Cosecant is the reciprocal of sine, so the sine piece we already have does most of the work. On $[- \frac{π}{2}, \frac{π}{2}]$ sine runs once through $[- 1, 1]$ . Taking reciprocals sends each non-zero value $s$ to $\frac{1}{s}$ , a number with $∣ \frac{1}{s} ∣ \geq 1$ , but we must drop $x = 0$ , where sine is zero. What is left, $[- \frac{π}{2}, \frac{π}{2}] - {0}$ , is carried one-one onto all numbers $y$ with $∣ y ∣ \geq 1$ , the set $𝐑 - (- 1, 1)$ , which is the whole range of cosecant.

Inverse cosecant

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

Secant is built from the cosine piece in the same way. Cosine runs once through $[- 1, 1]$ on $[0, π]$ and is zero at $\frac{π}{2}$ ; removing that point and taking reciprocals gives a one-one map of $[0, π] - {\frac{π}{2}}$ onto $𝐑 - (- 1, 1)$ .

Inverse secant

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

Each removed point turns into an asymptote of the inverse graph. As $∣ x ∣$ grows, ${cosec}^{- 1} x$ creeps towards $0$ and $\sec^{- 1} x$ creeps towards $\frac{π}{2}$ , but neither ever arrives (Fig. 2.5). Other branches come from sets such as $[\frac{π}{2}, \frac{3 π}{2}] - {π}$ for cosecant and $[π, 2 π] - {\frac{3 π}{2}}$ for secant.

4/4 $y = \sec^{- 1} x$ starts at $0$ when $x = 1$ and at $π$ when $x = - 1$ ; both arms approach $y = \frac{π}{2}$ .

Fig. 2.5Removed points become asymptotes after reflection.NCERT Fig. 2.3, 2.4 · p. 22

Equal scales from about minus 3.5 to 3.5 on both axes. Step 1 shows the two blue arms of y = cosec x between minus pi by 2 and pi by 2, one above y = 1 and one below y = minus 1, rising steeply near the y-axis. Step 2 adds the mirror line y = x and the red dashed image y = cosec inverse x: two arms for x at least 1 and x at most minus 1, starting at heights pi by 2 and minus pi by 2 and flattening towards the x-axis. Step 3 replaces them with the two arms of y = sec x on 0 to pi, with a dashed vertical asymptote at x = pi by 2. Step 4 adds the image y = sec inverse x: one arm rising from 0 at x = 1 and one falling from pi at x = minus 1, both approaching the dashed line y = pi by 2.

You rarely need new tables for these two. For $∣ x ∣ \geq 1$ , an angle whose secant is $x$ is an angle whose cosine is $\frac{1}{x}$ , and the principal ranges match, so $\sec^{- 1} x = \cos^{- 1} \frac{1}{x}$ . In the same way ${cosec}^{- 1} x = \sin^{- 1} \frac{1}{x}$ .

Inverse tangent and cotangent

Between two neighbouring asymptotes, tangent climbs through every real number exactly once. The pair nearest the origin, $x = \pm \frac{π}{2}$ , gives the principal piece $(- \frac{π}{2}, \frac{π}{2})$ , on which tangent is one-one and onto $𝐑$ . So $\tan^{- 1}$ accepts every real number, and its values stay strictly between $- \frac{π}{2}$ and $\frac{π}{2}$ .

Inverse tangent

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

Cotangent has its asymptotes at the multiples of $π$ , and between $0$ and $π$ it falls through every real number once. That interval, $(0, π)$ , is its principal piece.

Inverse cotangent

\cot^{- 1} : 𝐑 \to (0, π)

Both ranges are open intervals. No real $x$ has $\tan^{- 1} x = \frac{π}{2}$ , because $\tan \frac{π}{2}$ does not exist. The vertical asymptotes of the restricted tangent and cotangent become horizontal asymptotes of their inverses (Fig. 2.6). Other branches of $\tan^{- 1}$ use $(\frac{π}{2}, \frac{3 π}{2})$ , $(- \frac{3 π}{2}, - \frac{π}{2})$ and so on; other branches of $\cot^{- 1}$ use $(π, 2 π)$ , $(- π, 0)$ and so on.

2/4 $y = \tan^{- 1} x$ is defined for all $x$ and stays between the lines $y = \pm \frac{π}{2}$ .

Fig. 2.6Vertical asymptotes of the piece become horizontal asymptotes of the inverse.NCERT Fig. 2.5, 2.6 · p. 24

Equal scales from about minus 3.5 to 3.5. Step 1 shows the blue curve y = tan x between dashed vertical asymptotes at x = minus pi by 2 and pi by 2. Step 2 adds the mirror line y = x and the red dashed image y = tan inverse x, which runs across the whole picture and flattens towards dashed horizontal lines at y = pi by 2 and y = minus pi by 2. Step 3 shows instead y = cot x on 0 to pi, falling between asymptotes at x = 0 and x = pi. Step 4 adds y = cot inverse x, which falls from near pi on the left to near 0 on the right and crosses the y-axis at pi by 2.

All six principal branches

Table 2.2 puts the six choices side by side. Learn the ranges as two families: $\sin^{- 1}$ , $\tan^{- 1}$ and ${cosec}^{- 1}$ give angles around $0$ , inside $[- \frac{π}{2}, \frac{π}{2}]$ ; $\cos^{- 1}$ , $\cot^{- 1}$ and $\sec^{- 1}$ give angles inside $[0, π]$ .

Table 2.2Principal value branches of the inverse trigonometric functionsNCERT p. 25

Function	Domain	Principal range
$\sin^{- 1} x$	$[- 1, 1]$	$[- \frac{π}{2}, \frac{π}{2}]$
$\cos^{- 1} x$	$[- 1, 1]$	$[0, π]$
${cosec}^{- 1} x$	$𝐑 - (- 1, 1)$	$[- \frac{π}{2}, \frac{π}{2}] - {0}$
$\sec^{- 1} x$	$𝐑 - (- 1, 1)$	$[0, π] - {\frac{π}{2}}$
$\tan^{- 1} x$	$𝐑$	$(- \frac{π}{2}, \frac{π}{2})$
$\cot^{- 1} x$	$𝐑$	$(0, π)$

InteractiveBranch lab: six functions, three branches each

Buttons choose one of the six trigonometric functions and one of three branches: the principal one, or the next branch to the left or right. The graph shows the whole function as a faint dotted curve, the chosen branch in blue with its asymptotes, the mirror line y = x and, when the reflect box is ticked, the mirror image in red. A slider and buttons with exact values such as minus one half or root 3 set the input x; points P on the branch and P prime on the image move with it. A panel gives the angle on the chosen branch in radians, as a multiple of pi when it is one, and in degrees, compares it with the principal value, and says when an input such as one half has no inverse secant or cosecant.

Fig. 2.7Pick a function and a branch, reflect it in

y = x

, and move the input to read the inverse.

Use the lab to test the table. On the principal branch the readout always lands in the range listed above. Switch to a neighbouring branch and the same input produces a different angle with the same sine, cosine or tangent. Try an input of $\frac{1}{2}$ for secant: the lab refuses, because no angle has secant $\frac{1}{2}$ .

A unit circle makes the two families easy to picture (Fig. 2.8). The principal values of $\sin^{- 1}$ and $\tan^{- 1}$ are angles that end on the right half of the circle; those of $\cos^{- 1}$ and $\cot^{- 1}$ end on the upper half. To find a principal value, locate the angles with the given ratio and keep the one on the correct half.

3/4The line $x = - \frac{1}{2}$ meets the circle at $\frac{2 π}{3}$ and $- \frac{2 π}{3}$ ; the upper one is $\cos^{- 1} (- \frac{1}{2})$ .

Fig. 2.8Principal values of

\sin^{- 1}

and

\tan^{- 1}

end on the right half; those of

\cos^{- 1}

and

\cot^{- 1}

on the upper half.

A unit circle on the left and a notes panel on the right. Step 1 thickens the right half of the circle in blue, from minus pi by 2 at the bottom to pi by 2 at the top, and the panel lists the ranges of sin inverse and tan inverse. Step 2 thickens the upper half in orange, from 0 on the right to pi on the left, and the panel lists cos inverse and cot inverse. Step 3 draws the vertical line x = minus one half, which meets the circle at 2 pi by 3 above and minus 2 pi by 3 below; the upper point is filled and joined to the centre, so cos inverse of minus one half is 2 pi by 3. Step 4 draws the horizontal line y = minus one half, meeting the circle at minus pi by 6 on the right and 7 pi by 6 on the left; the right point is kept, so sin inverse of minus one half is minus pi by 6.

Definition

Principal value

The one answer an inverse trigonometric function gives once its principal branch is fixed: the angle with the given ratio that lies in that branch.

Example $\sin \frac{5 π}{6} = \frac{1}{2}$ , but the principal value of $\sin^{- 1} \frac{1}{2}$ is $\frac{π}{6}$ .

Example 1NCERT Example 1 · p. 26

Principal value of an inverse sine

Find the principal value of $\sin^{- 1} (\frac{\sqrt{3}}{2})$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 26

A negative input for inverse cotangent

Find the principal value of $\cot^{- 1} (- 1)$ .

Try it yourself first. Then reveal the solution one step at a time.

Example

Negative inputs for secant and cosecant

Find the principal values of (a) $\sec^{- 1} (- \sqrt{2})$ and (b) ${cosec}^{- 1} (- \frac{2}{\sqrt{3}})$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

Find the principal value of

\tan^{- 1} (- 1)

. Write it as

k π

and type

k

(for example, -1/3).

Show answer

Answer -0.25 π

\tan (- \frac{π}{4}) = - 1

and

- \frac{π}{4}

lies in

(- \frac{π}{2}, \frac{π}{2})

, so

k = - \frac{1}{4}

. The answer

\frac{3 π}{4}

also has tangent

- 1

, but it is outside the principal branch.

In Questions 1 to 12 of the practice set below, type an answer $k π$ as $k$ : for $- \frac{π}{3}$ type -1/3, and for $π$ type 1.

Practice · 14 questions

Practice 2.1: questions with answers

Short answerEasyNCERT Exercise 2.1, Q1
Find the principal value of $\sin^{- 1} (- \frac{\sqrt{3}}{2})$ .
π
Show solution
$\sin (- \frac{π}{3}) = - \sin \frac{π}{3} = - \frac{\sqrt{3}}{2}$ , and $- \frac{π}{3}$ lies in $[- \frac{π}{2}, \frac{π}{2}]$ . The principal value is $- \frac{π}{3}$ .
Short answerEasyNCERT Exercise 2.1, Q2
Find the principal value of $\cos^{- 1} (\frac{1}{\sqrt{2}})$ .
π
Show solution
$\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ and $\frac{π}{4} \in [0, π]$ , so the principal value is $\frac{π}{4}$ .
Short answerEasyNCERT Exercise 2.1, Q3
Find the principal value of ${cosec}^{- 1} (\frac{2}{\sqrt{3}})$ .
π
Show solution
$cosec y = \frac{2}{\sqrt{3}}$ means $\sin y = \frac{\sqrt{3}}{2}$ . With $y \in [- \frac{π}{2}, \frac{π}{2}] - {0}$ , $y = \frac{π}{3}$ .
Short answerEasyNCERT Exercise 2.1, Q4
Find the principal value of $\tan^{- 1} (- \frac{1}{\sqrt{3}})$ .
π
Show solution
$\tan (- \frac{π}{6}) = - \frac{1}{\sqrt{3}}$ and $- \frac{π}{6} \in (- \frac{π}{2}, \frac{π}{2})$ , so the principal value is $- \frac{π}{6}$ .
Short answerEasyNCERT Exercise 2.1, Q5
Find the principal value of $\cos^{- 1} (- \frac{\sqrt{3}}{2})$ .
π
Show solution
$\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ , so $\cos (π - \frac{π}{6}) = - \frac{\sqrt{3}}{2}$ . The angle $\frac{5 π}{6}$ lies in $[0, π]$ , so it is the principal value.
Short answerHardNCERT Exercise 2.1, Q6
Find the principal value of $\tan^{- 1} (\sqrt{3} - 2)$ .
π
Show solution
$\tan \frac{π}{12} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1)^{2}}{2} = 2 - \sqrt{3}$ . So $\tan (- \frac{π}{12}) = \sqrt{3} - 2$ , and $- \frac{π}{12}$ lies in $(- \frac{π}{2}, \frac{π}{2})$ . The principal value is $- \frac{π}{12}$ .
Short answerEasyNCERT Exercise 2.1, Q7
Find the principal value of $\sec^{- 1} (1)$ .
π
Show solution
$\sec 0 = 1$ and $0$ lies in $[0, π] - {\frac{π}{2}}$ , so $\sec^{- 1} 1 = 0$ . Compare ${cosec}^{- 1} 1 = \frac{π}{2}$ : it is the range of ${cosec}^{- 1}$ , not of $\sec^{- 1}$ , that leaves out $0$ .
Short answerEasyNCERT Exercise 2.1, Q8
Find the principal value of $\cot^{- 1} (\frac{1}{\sqrt{3}})$ .
π
Show solution
$\cot \frac{π}{3} = \frac{1}{\sqrt{3}}$ and $\frac{π}{3} \in (0, π)$ , so the principal value is $\frac{π}{3}$ .
Short answerEasyNCERT Exercise 2.1, Q9
Find the principal value of $\cos^{- 1} (- 1)$ .
π
Show solution
$\cos π = - 1$ and $π$ is the right-hand end of $[0, π]$ , so $\cos^{- 1} (- 1) = π$ .
Short answerEasyNCERT Exercise 2.1, Q10
Find the principal value of ${cosec}^{- 1} (- 2)$ .
π
Show solution
$cosec y = - 2$ means $\sin y = - \frac{1}{2}$ , and in $[- \frac{π}{2}, \frac{π}{2}] - {0}$ this gives $y = - \frac{π}{6}$ .
Short answerMediumNCERT Exercise 2.1, Q11
Find the value of $\cot^{- 1} (- 1) + \sin^{- 1} \frac{1}{2} - \sec^{- 1} \sqrt{2}$ .
π
Show solution
$\cot^{- 1} (- 1) = \frac{3 π}{4}$ , $\sin^{- 1} \frac{1}{2} = \frac{π}{6}$ and $\sec^{- 1} \sqrt{2} = \frac{π}{4}$ . The total is $\frac{9 π}{12} + \frac{2 π}{12} - \frac{3 π}{12} = \frac{8 π}{12} = \frac{2 π}{3}$ .
Short answerMediumNCERT Exercise 2.1, Q12
Find the value of $2 \cos^{- 1} \frac{\sqrt{3}}{2} + \sin^{- 1} (- 1)$ .
π
Show solution
$\cos^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{6}$ and $\sin^{- 1} (- 1) = - \frac{π}{2}$ , so the value is $\frac{π}{3} - \frac{π}{2} = - \frac{π}{6}$ .
Quick answerEasyNCERT Exercise 2.1, Q13
Which statement about $y = \cot^{- 1} x$ is true for every real $x$ ?
$y = 0$ once $x$ is large enough $- \frac{π}{2} < y < \frac{π}{2}$ $0 < y < π$ $y = \frac{π}{2}$ when $x = 1$
Show answer
Answer (C) $0 < y < π$
The principal branch of $\cot^{- 1}$ is the open interval $(0, π)$ : $\cot 0$ and $\cot π$ do not exist, so $y$ gets close to $0$ for large $x$ but never equals it. $(- \frac{π}{2}, \frac{π}{2})$ is the range of $\tan^{- 1}$ , and $\cot^{- 1} 1 = \frac{π}{4}$ .
Quick answerMediumNCERT Exercise 2.1, Q14
$\sec^{- 1} (- \sqrt{2}) - \tan^{- 1} (- \frac{1}{\sqrt{3}})$ is equal to
$\frac{7 π}{12}$ $\frac{11 π}{12}$ $- \frac{π}{12}$ $- \frac{5 π}{12}$
Show answer
Answer (B) $\frac{11 π}{12}$
$\sec^{- 1} (- \sqrt{2}) = \frac{3 π}{4}$ and $\tan^{- 1} (- \frac{1}{\sqrt{3}}) = - \frac{π}{6}$ , so the value is $\frac{3 π}{4} + \frac{π}{6} = \frac{11 π}{12}$ . Taking $\sec^{- 1} (- \sqrt{2}) = - \frac{π}{4}$ , outside its range, leads to $- \frac{π}{12}$ .

2.3 Properties of Inverse Trigonometric Functions

Every identity in this section has fine print. It holds only for those $x$ where both sides are defined and the angles involved stay inside the principal branches. Ignoring the fine print is the commonest source of wrong answers in this chapter, so each result below comes with its interval.

PropertyUndoing a trigonometric function

$\sin (\sin^{- 1} x) = x$ for $x \in [- 1, 1]$ , and $\sin^{- 1} (\sin x) = x$ for $x \in [- \frac{π}{2}, \frac{π}{2}]$ . Likewise $\cos (\cos^{- 1} x) = x$ for $x \in [- 1, 1]$ and $\cos^{- 1} (\cos x) = x$ for $x \in [0, π]$ ; $\tan (\tan^{- 1} x) = x$ for every real $x$ and $\tan^{- 1} (\tan x) = x$ for $x \in (- \frac{π}{2}, \frac{π}{2})$ . The other three functions obey the same pattern on their own branches.

The two halves are not alike. The first, $\sin (\sin^{- 1} x) = x$ , holds on the whole domain of $\sin^{- 1}$ . The second breaks as soon as $x$ leaves $[- \frac{π}{2}, \frac{π}{2}]$ : for instance $\sin^{- 1} (\sin π) = \sin^{- 1} 0 = 0$ , not $π$ . The graph of $y = \sin^{- 1} (\sin x)$ shows what happens instead (Fig. 2.9).

3/3At $x = \frac{5 π}{4}$ the graph gives $π - \frac{5 π}{4} = - \frac{π}{4}$ .

Fig. 2.9

\sin^{- 1} (\sin x)

equals

x

only on the principal branch; elsewhere it folds back.

Equal scales, x from about minus pi to 2 pi. Step 1 shades the strip from minus pi by 2 to pi by 2, where a short blue segment lies exactly on the dashed line y = x. Step 2 adds the rest of the blue graph: a zigzag with slope minus 1 from pi by 2 down to minus pi by 2 at 3 pi by 2, then slope 1 again, and a matching fold to the left, never going above pi by 2 or below minus pi by 2. Step 3 marks the point above x = 5 pi by 4, which lies at height minus pi by 4, with dotted guides to both axes.

The zigzag has slope $1$ on $[- \frac{π}{2}, \frac{π}{2}]$ , slope $- 1$ on $[\frac{π}{2}, \frac{3 π}{2}]$ , slope $1$ again on $[\frac{3 π}{2}, \frac{5 π}{2}]$ , and repeats with period $2 π$ . To evaluate $\sin^{- 1} (\sin a)$ for any $a$ , replace $a$ by an angle in $[- \frac{π}{2}, \frac{π}{2}]$ with the same sine, using $\sin (π - a) = \sin a$ and $\sin (a + 2 π) = \sin a$ . The same idea works for the other functions with their own symmetries, such as $\cos (2 π - a) = \cos a$ and $\tan (a + π) = \tan a$ .

Most simplifications follow one recipe. Put $x$ equal to a trigonometric function of a new angle $θ$ , rewrite the expression with a standard identity, and use the property above to cancel the inverse. The step that needs care is the last: before cancelling, check that the angle you have reached lies in the right principal branch.

Example 3NCERT Example 3 · p. 28

One expression, two answers

Show that $\cos^{- 1} (2 x^{2} - 1) = 2 \cos^{- 1} x$ for $0 \leq x \leq 1$ , and that $\cos^{- 1} (2 x^{2} - 1) = 2 π - 2 \cos^{- 1} x$ for $- 1 \leq x \leq 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Fig. 2.10 compares the two sides of Example 3. They coincide on $[0, 1]$ . On $[- 1, 0)$ the curve $y = 2 \cos^{- 1} x$ climbs above $π$ , a height that no value of $\cos^{- 1}$ can reach, so the left-hand side has to fold back.

3/3They agree on $[0, 1]$ . At $x = - \frac{1}{2}$ they give $\frac{4 π}{3}$ and $\frac{2 π}{3}$ , and $\frac{2 π}{3} = 2 π - \frac{4 π}{3}$ .

Fig. 2.10The two sides of Example 3 agree only for

0 \leq x \leq 1

The x-axis runs from minus 1 to 1 and the y-axis from 0 to 2 pi. Step 1 shows the blue curve y = 2 cos inverse x, falling from 2 pi at x = minus 1 to 0 at x = 1. Step 2 adds the orange dashed curve y = cos inverse of (2 x squared minus 1), which rises from 0 at x = minus 1 to pi at x = 0 and falls back to 0, symmetric about the y-axis. Step 3 shades the part from 0 to 1, labelled equal here, where the two curves coincide, and marks x = minus one half, where the blue curve is at 4 pi by 3 and the orange one at 2 pi by 3.

Example 4NCERT Example 4 · p. 28

Sum-to-product inside an inverse tangent

Write $\tan^{- 1} (\frac{\sin 3 x + \sin x}{\cos 3 x + \cos x})$ , $- \frac{π}{4} < x < \frac{π}{4}$ , in the simplest form.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 29

A surd that hides an angle

Write $\sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}})$ , $x > 0$ , in the simplest form.

Try it yourself first. Then reveal the solution one step at a time.

Miscellaneous examples

Example 6NCERT Example 6 · p. 30

An angle outside the branch

Find $\sin^{- 1} (\sin \frac{5 π}{4})$ .

Try it yourself first. Then reveal the solution one step at a time.

Compositions such as $\cos (\tan^{- 1} x)$ or $\tan (\sin^{- 1} \frac{7}{25})$ are handled with a right triangle. Name the inner angle $θ$ , draw a triangle whose sides fit the given ratio, find the third side by Pythagoras and read off the ratio you need (Fig. 2.11). For a negative input the triangle gives the size of the answer, and the quadrant of $θ$ gives its sign.

3/3Now $\cos θ = \frac{24}{25}$ and $\tan θ = \frac{7}{24}$ , so $\tan (\sin^{- 1} \frac{7}{25}) = \frac{7}{24}$ .

Fig. 2.11Two sides from the ratio, the third by Pythagoras, then any ratio you like.

A right triangle with the angle theta at the left corner and the right angle at the bottom right. A panel at the top says theta equals sin inverse of 7 by 25. Step 1 labels the short vertical side opposite theta as 7 and the slanting hypotenuse as 25, with a question mark on the base. Step 2 colours the base orange and labels it 24, and the panel shows the square root of 25 squared minus 7 squared equals 24. Step 3 names the sides adjacent and opposite, and the panel reads cos theta equals 24 by 25 and tan theta equals 7 by 24.

Example

Watch the quadrant

Find (a) $\tan (\cos^{- 1} (- \frac{3}{5}))$ and (b) $\cos (\tan^{- 1} (- 2))$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Find

\cos^{- 1} (\cos \frac{7 π}{4})

. Write it as

k π

and type

k

Show answer

Answer 0.25 π

\cos \frac{7 π}{4} = \cos (2 π - \frac{7 π}{4}) = \cos \frac{π}{4}

, and

\frac{π}{4} \in [0, π]

. So the value is

\frac{π}{4}

and

k = \frac{1}{4}

From historyNCERT Historical Note · p. 33

From jyā to arc sine

The sine function grew out of Indian astronomy. The siddhānta astronomers worked with the half-chord of a circle, which they called jyā or ardha-jyā, rather than with the whole chord used by Greek astronomers. Āryabhaṭa (born 476 CE) gave a table of these half-chords at steps of $3 {\frac{3}{4}}^{\circ}$ in his Āryabhaṭīya (499 CE). The word travelled into Arabic as jība, was later read as jaib (a fold or pocket) and became the Latin sinus: our “sine”.

Later astronomers sharpened these tools. Bhāskara I (7th century) found a simple fraction that approximates the sine of an angle remarkably well, Brahmagupta gave rules for reading between the entries of a sine table, and Bhāskara II (born 1114 CE) worked out exact values such as $\sin 18^{\circ}$ and $\sin 36^{\circ}$ .

The first inverse ideas came from Kerala. Mādhava of Saṅgamagrāma (about 1340 to 1425) found an endless series for an arc in terms of its tangent, which we now write as $\tan^{- 1} x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \dots$ , and from it a series for $π$ . The Malayalam text Yuktibhāṣā (about 1530) set out proofs of these results, along with the expansion of $\sin (A + B)$ .

The notation $\sin^{- 1} x$ was proposed by the English astronomer John Herschel in 1813; the older name arc sine records that the answer is an arc of the unit circle. Much earlier, Thales of Miletus (about 600 BCE) is said to have found the height of a pyramid by comparing its shadow with the shadow of an upright staff, an early use of the ratio we now call the tangent of the sun's altitude. The Āryabhaṭīya itself poses shadow problems of the same kind, about a gnomon (śaṅku) standing near a lamp.

Timeline

Milestones on the way to inverse trigonometric functions

499 CE
Āryabhaṭīya
A table of half-chords (jyā) at steps of $3 {\frac{3}{4}}^{\circ}$ .
c. 629 CE
Bhāskara I
A simple rational formula that approximates the sine.
665 CE
Brahmagupta
Second-order interpolation for sine tables.
1150 CE
Bhāskara II
Exact values such as $\sin 18^{\circ}$ and $\sin 36^{\circ}$ .
c. 1400
Mādhava
Series for the arc tangent, $\tan^{- 1} x$ , and for $π$ .
c. 1530
Yuktibhāṣā
Proofs of the Kerala series, written in Malayalam.
1813
John Herschel
Introduces the notation $\sin^{- 1} x$ for the arc sine.

Flash cards · 8

Principal branches: quick recall

1 / 8

Recap

Inverse Trigonometric Functions: key points and summary

A trigonometric function is many-one on its natural domain, so it has an inverse only after its domain is restricted to an interval where it is one-one and onto (Section 2.1).
If $f$ is one-one and onto, $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (y)) = y$ , and the graph of $f^{- 1}$ is the mirror image of the graph of $f$ in the line $y = x$ (Section 2.2).
Each admissible interval gives a branch of the inverse. The agreed one is the principal value branch, and a value taken from it is a principal value (Inverse sine).
$\sin^{- 1} : [- 1, 1] \to [- \frac{π}{2}, \frac{π}{2}]$ and $\cos^{- 1} : [- 1, 1] \to [0, π]$ (Inverse cosine).
${cosec}^{- 1}$ and $\sec^{- 1}$ are defined for $∣ x ∣ \geq 1$ , with ranges $[- \frac{π}{2}, \frac{π}{2}] - {0}$ and $[0, π] - {\frac{π}{2}}$ (Inverse cosecant and secant).
$\tan^{- 1}$ and $\cot^{- 1}$ are defined for every real $x$ , with open ranges $(- \frac{π}{2}, \frac{π}{2})$ and $(0, π)$ (Inverse tangent and cotangent).
$\sin^{- 1} x$ is an angle, while $(\sin x)^{- 1} = cosec x$ is a reciprocal. With no branch named, the principal branch is meant (All six principal branches).
For a negative input, $\sin^{- 1}$ , $\tan^{- 1}$ and ${cosec}^{- 1}$ give a negative angle, while $\cos^{- 1}$ , $\cot^{- 1}$ and $\sec^{- 1}$ give an angle between $\frac{π}{2}$ and $π$ (negative inputs).
$\sin (\sin^{- 1} x) = x$ on $[- 1, 1]$ , but $\sin^{- 1} (\sin x) = x$ only on $[- \frac{π}{2}, \frac{π}{2}]$ ; each other function obeys the same pair of rules on its own branch (Property).
To find $\sin^{- 1} (\sin a)$ for $a$ outside the branch, replace $a$ by the angle in $[- \frac{π}{2}, \frac{π}{2}]$ that has the same sine (Example 6).
Substitutions such as $x = \sin θ$ , $\cos θ$ or $\tan θ$ simplify inverse trigonometric expressions; the interval of $θ$ decides the final form (Example 3).
Compositions such as $\tan (\cos^{- 1} x)$ are read from a right triangle, with the sign fixed by the quadrant of the principal value (Fig. 2.11).

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 2.1, Q1	Practice 2.1 · Q1
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NCERT Exercise 2.1, Q3	Practice 2.1 · Q3
NCERT Exercise 2.1, Q4	Practice 2.1 · Q4
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NCERT Exercise 2.1, Q9	Practice 2.1 · Q9
NCERT Exercise 2.1, Q10	Practice 2.1 · Q10
NCERT Exercise 2.1, Q11	Practice 2.1 · Q11
NCERT Exercise 2.1, Q12	Practice 2.1 · Q12
NCERT Exercise 2.1, Q13	Practice 2.1 · Q13
NCERT Exercise 2.1, Q14	Practice 2.1 · Q14
NCERT Exercise 2.2, Q1	Practice 2.2 · Q1
NCERT Exercise 2.2, Q2	Practice 2.2 · Q2
NCERT Exercise 2.2, Q3	Practice 2.2 · Q3
NCERT Exercise 2.2, Q4	Practice 2.2 · Q4
NCERT Exercise 2.2, Q5	Practice 2.2 · Q5
NCERT Exercise 2.2, Q6	Practice 2.2 · Q6
NCERT Exercise 2.2, Q7	Practice 2.2 · Q7
NCERT Exercise 2.2, Q8	Practice 2.2 · Q8
NCERT Exercise 2.2, Q9	Practice 2.2 · Q9
NCERT Exercise 2.2, Q10	Practice 2.2 · Q10
NCERT Exercise 2.2, Q11	Practice 2.2 · Q11
NCERT Exercise 2.2, Q12	Practice 2.2 · Q12
NCERT Exercise 2.2, Q13	Practice 2.2 · Q13
NCERT Exercise 2.2, Q14	Practice 2.2 · Q14
NCERT Exercise 2.2, Q15	Practice 2.2 · Q15
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises on Chapter 2 · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises on Chapter 2 · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises on Chapter 2 · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises on Chapter 2 · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises on Chapter 2 · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises on Chapter 2 · Q6
NCERT Miscellaneous Exercise, Q7	Miscellaneous exercises on Chapter 2 · Q7
NCERT Miscellaneous Exercise, Q8	Miscellaneous exercises on Chapter 2 · Q8
NCERT Miscellaneous Exercise, Q9	Miscellaneous exercises on Chapter 2 · Q9
NCERT Miscellaneous Exercise, Q10	Miscellaneous exercises on Chapter 2 · Q10
NCERT Miscellaneous Exercise, Q11	Miscellaneous exercises on Chapter 2 · Q11
NCERT Miscellaneous Exercise, Q12	Miscellaneous exercises on Chapter 2 · Q12
NCERT Miscellaneous Exercise, Q13	Miscellaneous exercises on Chapter 2 · Q13
NCERT Miscellaneous Exercise, Q14	Miscellaneous exercises on Chapter 2 · Q14

Quick check: MCQs on Inverse Trigonometric Functions

2.1 Introduction2 questions
Inverse cosine3 questions
All six principal branches4 questions
2.3 Properties of Inverse Trigonometric Functions4 questions
Miscellaneous examples3 questions

Glossary · 9

Key terms (English–Hindi)

asymptoteअनंतस्पर्शीanantasparshi: A line that a graph approaches ever more closely without reaching it, such as $y = \frac{π}{2}$ for $y = \tan^{- 1} x$ .First seen in the chapter →

branchशाखाshaakhaa: One interval on which a trigonometric function is one-one and onto its range, or the inverse it gives.First seen in the chapter →

inverse functionप्रतिलोम फलनpratilom phalan: For a bijection $f : X \to Y$ , the function $f^{- 1} : Y \to X$ that sends each $f (x)$ back to $x$ .First seen in the chapter →
inverse trigonometric functionप्रतिलोम त्रिकोणमितीय फलनpratilom trikonmitiya phalan: The inverse of a trigonometric function restricted to one branch; for example $\sin^{- 1} x$ is the angle in $[- \frac{π}{2}, \frac{π}{2}]$ whose sine is $x$ .First seen in the chapter →

many-one functionबहुएक फलनbahu-ek phalan: A function that sends two or more different inputs to the same output, so it is not one-one.First seen in the chapter →

one-one and onto functionएकैकी आच्छादक फलनekaiki aachchhaadak phalan: A function that is both one-one and onto; exactly these functions have inverses.First seen in the chapter →

principal valueमुख्य मानmukhya maan: The angle an inverse trigonometric function returns from its principal value branch, such as $\frac{π}{6}$ for $\sin^{- 1} \frac{1}{2}$ .First seen in the chapter →
principal value branchमुख्य मान शाखाmukhya maan shaakhaa: The branch fixed by convention as the range of an inverse trigonometric function, such as $[0, π]$ for $\cos^{- 1}$ .First seen in the chapter →

restricted domainप्रतिबंधित प्रांतpratibandhit praant: A part of a function's natural domain, chosen so that the function becomes one-one and onto its range.First seen in the chapter →

Inverse Trigonometric Functions: Class 12 Maths · Chapter 2

What you will learn in Inverse Trigonometric Functions

2.1 Introduction

2.2 Basic Concepts

Inverse sine

Inverse cosine

Inverse cosecant and secant

Inverse tangent and cotangent

All six principal branches

Practice 2.1: questions with answers

2.3 Properties of Inverse Trigonometric Functions

Practice 2.2: questions with answers

Miscellaneous examples

Inverse Trigonometric Functions: key points and summary

Practice questions with answers (NCERT “Miscellaneous exercises on Chapter 2”)

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Quick check: MCQs on Inverse Trigonometric Functions

Key terms (English–Hindi)

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