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Why the inverse of sine needs a rule first
Chapter 1 defined when a function is invertible: exactly when it is one-one and onto, so a reverse rule exists and is unambiguous. This chapter asks the same question of $\sin$, $\cos$, and $\tan$, and the answer is not immediately yes.
None of the three trig functions is one-one over its full natural domain. $\sin(0) = \sin(\pi) = \sin(2 \pi) = 0$ — three different inputs, one output, repeated infinitely often as the angle keeps turning, and the same failure holds for $\cos$ and $\tan$. An inverse cannot be defined on a function that keeps repeating its own outputs, so “the inverse of $\sin$” is not yet a well-defined question.
The fix is the one Chapter 1 already names: restrict the domain to a piece where the function is one-one, and, chosen well, onto its full range too. Such a piece is called, by convention, the PRINCIPAL BRANCH. This chapter fixes one principal branch for each of the six trig functions and defines the inverse only on that branch — never on the function’s full natural domain.
The principal branches, and their graphs
Restrict $\sin$ to the interval $[-\pi/2, \pi/2]$ and it becomes one-one, matching every angle in that interval to a distinct value in $[-1, 1]$ — and onto $[-1, 1]$ as well, since every value in that range is hit somewhere on the restricted piece. This is the piece Chapter 2 fixes as sine’s principal branch, and the inverse is defined only on it.
$\sin^{-1}: [-1, 1] \to [-\pi/2, \pi/2]$ sends each $y$ back to the one $x$ in $[-\pi/2, \pi/2]$ with $\sin x = y$. $\sin^{-1}(1) = \pi/2$, because $\sin(\pi/2) = 1$ and $\pi/2$ lies in the branch. $\sin^{-1}(0) = 0$, for the same reason: $\sin(0) = 0$, and $0$ sits inside $[-\pi/2, \pi/2]$.
Every other angle with the same sine, outside this branch, is simply not a candidate answer. The branch is not a restriction imposed after the fact; it is what makes the inverse exist at all.
Cosine’s principal branch is not the same interval as sine’s. Restricted to $[0, \pi]$, $\cos$ is one-one and onto $[-1, 1]$ — a branch running from $0$ to $\pi$, not symmetric about $0$ the way sine’s is.
$\cos^{-1}: [-1, 1] \to [0, \pi]$ sends each $y$ back to the one $x$ in $[0, \pi]$ with $\cos x = y$. $\cos^{-1}(1) = 0$, since $\cos(0) = 1$ and $0$ is inside $[0, \pi]$. $\cos^{-1}(-1) = \pi$, since $\cos(\pi) = -1$ and $\pi$ sits at the far end of the same branch.
Each inverse trig function earns its own principal branch, chosen for that function specifically — there is no single interval that works for all three.
Tangent’s principal branch is open at both ends. Restricted to $(-\pi/2, \pi/2)$ — excluding the two endpoints, where $\tan$ is undefined — $\tan$ is one-one and onto all of $\mathbb{R}$.
$\tan^{-1}: \mathbb{R} \to (-\pi/2, \pi/2)$ sends each real $y$ back to the one $x$ in $(-\pi/2, \pi/2)$ with $\tan x = y$. $\tan^{-1}(0) = 0$, since $\tan(0) = 0$. $\tan^{-1}(1) = \pi/4$, since $\tan(\pi/4) = 1$ and $\pi/4$ lies inside the branch.
Unlike sine and cosine, tangent’s inverse takes every real number as an input — its restricted branch is already onto the whole real line, with nothing left over.
Cotangent, secant, and cosecant inverse follow the same pattern: each gets a branch on which the original function is one-one, and the inverse is defined only there. $\cot^{-1}: \mathbb{R} \to (0, \pi)$, mirroring tangent’s role but never touching $0$ or $\pi$ themselves.
$\sec^{-1}$ and $\csc^{-1}$ both exclude the interval $(-1, 1)$ from their domain — no input strictly between $-1$ and $1$ is ever hit, since neither $\sec$ nor $\csc$ ever takes a value in that range. $\sec^{-1}$ has range $[0, \pi] - {\pi/2}$; $\csc^{-1}$ has range $[-\pi/2, \pi/2] - {0}$, each excluding the one point where the original function is undefined.
NCERT gives these three far less weight than sine, cosine, and tangent inverse, because most problems reduce them back to the first three by a reciprocal identity — $\sec^{-1} x = \cos^{-1}(1/x)$ for $|x| \geq 1$, for instance.
A function and its inverse are always mirror images of each other across the line $y = x$ — swapping $x$ and $y$ is exactly what taking an inverse does to every point on a graph. That general fact, from Chapter 1, applies here to three familiar curves.
The graph of $\sin^{-1} x$ rises from $(-1, -\pi/2)$ through the origin to $(1, \pi/2)$, strictly increasing throughout — the mirror of sine’s own rising piece on $[-\pi/2, \pi/2]$. The graph of $\cos^{-1} x$ falls from $(-1, \pi)$ to $(1, 0)$, strictly decreasing throughout, mirroring cosine’s falling piece on $[0, \pi]$.
*$\tan^{-1} x$ behaves differently from the other two: it is strictly increasing on all of $\mathbb{R}$, but it never reaches $\pi/2$ or $-\pi/2$* — those become the horizontal asymptotes the curve flattens toward as $x \to -\infty$ and $x \to \infty$, without ever touching either one.
Evaluating principal values
Evaluate $\sin^{-1}(1/2)$
- $\sin^{-1}(1/2)$
Find the angle $x \in [-\pi/2, \pi/2]$ with $\sin x = 1/2$. - $\sin(\pi/6) = 1/2$
A standard value, and $\pi/6$ lies inside the principal branch $[-\pi/2, \pi/2]$. - *$\sin^{-1}(1/2) = \pi/6$*
The unique angle in the branch with sine $1/2$ — the only valid answer. - $\sin(5\pi/6) = 1/2$ too, but $5\pi/6$ lies outside $[-\pi/2, \pi/2]$
A second angle shares the same sine value, yet it is not a candidate — only angles inside the principal branch count.
Evaluate $\cos^{-1}(-1/\sqrt{2})$
- $\cos^{-1}(-1/\sqrt{2})$
Find the angle $x \in [0, \pi]$ with $\cos x = -1/\sqrt{2}$. - $\cos(\pi/4) = 1/\sqrt{2}$
A standard value; cosine is positive here, not the sign needed. - cosine is negative in the second quadrant, so $\cos(\pi - \pi/4) = -\cos(\pi/4) = -1/\sqrt{2}$
The supplementary angle flips the sign, giving the negative value required. - $\pi - \pi/4 = 3\pi/4$, and $3\pi/4 \in [0, \pi]$
The resulting angle lies inside the principal branch for cosine inverse. - *$\cos^{-1}(-1/\sqrt{2}) = 3\pi/4$*
The unique angle in $[0, \pi]$ with cosine $-1/\sqrt{2}$.
Evaluate $\tan^{-1}(-1)$
- $\tan^{-1}(-1)$
Find the angle $x \in (-\pi/2, \pi/2)$ with $\tan x = -1$. - $\tan(\pi/4) = 1$
A standard value, positive — the sign still needs fixing. - $\tan$ is an odd function, so $\tan(-\pi/4) = -\tan(\pi/4) = -1$
Odd symmetry flips the sign without leaving the branch. - $-\pi/4 \in (-\pi/2, \pi/2)$
The resulting angle lies inside the principal branch for tangent inverse. - *$\tan^{-1}(-1) = -\pi/4$*
The unique angle in the branch with tangent $-1$.
When sine inverse of sine is not x
It is tempting to treat $\sin^{-1}(\sin x) = x$ as automatic for every real $x$ — an inverse is supposed to undo the function it inverts. *That cancellation only holds when $x$ is already inside the principal branch $[-\pi/2, \pi/2]$.*
Outside the branch, $\sin^{-1}$ still has to return a value inside $[-\pi/2, \pi/2]$ — that is what the function is defined to do — so it cannot hand back an $x$ that lives outside it. What comes back instead is the angle inside $[-\pi/2, \pi/2]$ that shares the same sine as $x$, not $x$ itself.
The same caution carries over to the other two: $\cos^{-1}(\cos x) = x$ only for $x \in [0, \pi]$, and $\tan^{-1}(\tan x) = x$ only for $x \in (-\pi/2, \pi/2)$. Every inverse trig identity of this shape carries a domain condition, and skipping it is how a correct-looking simplification goes wrong.
Evaluate $\sin^{-1}(\sin(3\pi/4))$
- $\sin^{-1}(\sin(3\pi/4))$
$3\pi/4$ does not lie in the principal branch $[-\pi/2, \pi/2]$, so the answer is not simply $3\pi/4$. - $\sin(3\pi/4) = \sin(\pi - \pi/4) = \sin(\pi/4) = 1/\sqrt{2}$
Reduce the angle using the supplementary-angle identity for sine first. - $\sin^{-1}(1/\sqrt{2}) = \pi/4$
Apply the definition to the reduced value; $\pi/4$ lies inside the principal branch. - *$\sin^{-1}(\sin(3\pi/4)) = \pi/4$*
The branch angle sharing the same sine as $3\pi/4$, not $3\pi/4$ itself.
Two different meanings for the same exponent
The notation $\sin^{-1} x$ invites a reflex: a $-1$ exponent usually means reciprocal, so surely $\sin^{-1} x$ means $1/(\sin x)$. *It does not — $\sin^{-1} x$ is function notation for the inverse sine function, an angle, while $1/(\sin x)$ is a number, and that number already has its own name: $\csc x$.*
The two are almost never equal. $\sin^{-1}(1/2) = \pi/6 \approx 0.52$, an angle in radians. $1/(\sin(1/2)) \approx 2.09$, a plain ratio computed from the angle $1/2$ radian, not from a “power” of the value $1/2$. Different inputs, different kinds of output.
The $-1$ in $\sin^{-1}$ is a labelling convention borrowed from function-inverse notation generally — the same $-1$ that names $f^{-1}$ for any invertible $f$ — not the arithmetic exponent it looks like. Reading it as “raise to the power $-1$” is the single most common way this notation gets misread.
Complementary angle identities
Three pairs of inverse trig functions add to a fixed constant, $\pi/2$, each on its own shared domain. $\sin^{-1} x + \cos^{-1} x = \pi/2$ for every $x \in [-1, 1]$. $\tan^{-1} x + \cot^{-1} x = \pi/2$ for every $x \in \mathbb{R}$. $\sec^{-1} x + \csc^{-1} x = \pi/2$ for every $x$ with $|x| \geq 1$.
Each identity is the co-function relationship from Class 11, carried over unchanged: $\cos((\pi/2) - \theta) = \sin \theta$ says sine and cosine of complementary angles match, and the same pairing holds for tangent-cotangent and secant-cosecant. Restricting to the principal branches does not break the relationship — it only fixes which angle each side of the identity names.
These three constants are worth memorising as a set: given one member of a pair, the other follows by subtracting from $\pi/2$, without recomputing anything.
Check $\sin^{-1} x + \cos^{-1} x = \pi/2$ at $x = 1/2$
- $\sin^{-1}(1/2) = \pi/6$
Already known from evaluating sine inverse at $1/2$. - $\cos^{-1}(1/2) = \pi/3$
$\cos(\pi/3) = 1/2$, and $\pi/3$ lies in the principal branch $[0, \pi]$. - $\pi/6 + \pi/3 = \pi/6 + 2\pi/6 = 3\pi/6 = \pi/2$
Add the two values, converting to a common denominator first. - *$\sin^{-1}(1/2) + \cos^{-1}(1/2) = \pi/2$*
Confirms the complementary identity at this one value.
Negating the input, two different ways
Negating the input to an inverse trig function does not always negate the output — it depends on which branch the function uses. Two patterns cover all six functions.
For $\sin^{-1}$, $\tan^{-1}$, and $\csc^{-1}$ — the three whose principal branch is symmetric about $0$ — negating the input negates the output outright: $\sin^{-1}(-x) = -\sin^{-1} x$, $\tan^{-1}(-x) = -\tan^{-1} x$, $\csc^{-1}(-x) = -\csc^{-1} x$.
*For $\cos^{-1}$, $\cot^{-1}$, and $\sec^{-1}$ — whose branches run from $0$, not centred on it — negating the input instead subtracts the output from $\pi$*: $\cos^{-1}(-x) = \pi - \cos^{-1} x$, $\cot^{-1}(-x) = \pi - \cot^{-1} x$, $\sec^{-1}(-x) = \pi - \sec^{-1} x$. The shape of the branch, not the function’s formula, decides which pattern applies.
The addition formula for tangent inverse
Two inverse tangents add to a third, under one condition. For $x, y$ with $x y < 1$: $\tan^{-1} x + \tan^{-1} y = \tan^{-1}((x + y)/(1 - x y))$.
This is the familiar tangent addition formula from Class 11, re-expressed for inverses. *The condition $x y < 1$ is not optional bookkeeping — when $x y > 1$, the plain formula lands the sum in the wrong branch, and an extra $+ \pi$ or $- \pi$ term is needed to correct it.*
NCERT restricts this identity to the $x y < 1$ case for exactly that reason: the formula as stated is only guaranteed correct within it.
Find $\tan^{-1}(1/2) + \tan^{-1}(1/3)$
- $x = 1/2$, $y = 1/3$
Identify both values before checking the condition. - $x y = 1/6 < 1$
The condition for the plain addition formula holds, so it applies directly. - $(x + y)/(1 - x y) = (1/2 + 1/3)/(1 - 1/6) = (5/6)/(5/6) = 1$
Substitute into the formula and simplify. - *$\tan^{-1}(1/2) + \tan^{-1}(1/3) = \tan^{-1}(1) = \pi/4$*
$\tan^{-1}(1) = \pi/4$ is a standard value.
A forward look — inverse trig functions in integrals
Inverse trig functions reappear later in the course, in a role that has nothing to do with angles at first glance: as antiderivatives.
*Chapter 7 will show that the derivative of $\sin^{-1} x$ is $1/\sqrt{1 - x^2}$ — which means integrating $1/\sqrt{1 - x^2}$ gives back $\sin^{-1} x + C$.* No integration technique is needed yet; nothing here asks for one.
This is only a name worth recognising when it resurfaces: a fraction with a square root of $1 - x^2$ in the denominator is a strong hint that an inverse trig function is hiding behind it.
Restriction, and checking the branch first
Two ideas run through this chapter, both about precision under restriction.
First, “the inverse of $\sin$” only means something once a principal branch is fixed — $[-\pi/2, \pi/2]$ for sine, $[0, \pi]$ for cosine, $(-\pi/2, \pi/2)$ for tangent — the same bijective-iff-invertible idea from Chapter 1, applied concretely to three specific curves.
Second, every identity in this chapter carries a domain condition of its own — $\sin^{-1}(\sin x) = x$ only inside the branch, the complementary pairs on their shared domain, the addition formula only for $x y < 1$ — and skipping that check is the single most common way to turn a right-looking answer wrong.
Both habits come down to the same discipline: fix the branch first, then check every condition an identity carries — never trust an answer that looks right by shape alone.
Practice set
Exercise 2.1 practises finding principal values directly. Evaluate $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, and the other three inverse trig functions at standard values like $1/2$, $1/\sqrt{2}$, and $\sqrt{3}/2$, and at their negatives — always reporting the answer from the principal branch, not just any angle with the right ratio.
- practice Evaluate $\sin^{-1}(1/2)$.
- practice Evaluate $\cos^{-1}(1)$.
- practice Evaluate $\tan^{-1}(-1)$.
- practice Evaluate $\sin^{-1}(-1)$.
- practice Evaluate $\cos^{-1}(-1/2)$.
- practice Evaluate $\tan^{-1}(\sqrt{3})$.
- practice Evaluate $\cot^{-1}(1)$.
- practice Evaluate $\sec^{-1}(-\sqrt{2})$.
- practice Evaluate $\csc^{-1}(-2)$.
- practice Evaluate $\cot^{-1}(-\sqrt{3})$.
- practice Evaluate $\sin^{-1}(\sqrt{3}/2) + \cos^{-1}(1/2)$.
- practice Which of these equals $\tan^{-1}(1/\sqrt{3})$?
- practice Evaluate $\cos^{-1}(1/2) - \sin^{-1}(-1/2)$.
- practice Evaluate $\sec^{-1}(2) + \csc^{-1}(2)$.
- practice Which of these equals $\sin^{-1}(-1) + \cos^{-1}(-1)$?
Answers
- $\pi/6$
- $0$
- $-\pi/4$
- $-\pi/2$
- $2 \pi/3$
- $\pi/3$
- $\pi/4$
- $3 \pi/4$
- $-\pi/6$
- $5 \pi/6$
- $2 \pi/3$
- A — $\pi/6$.
- $\pi/2$
- $\pi/2$
- C — $\pi/2$.
Exercise 2.2 practises the elementary properties built up across the chapter. It covers simplifying expressions like $\sin^{-1}(\sin x)$ for $x$ outside the principal branch, applying the complementary-angle identities, and combining two inverse tangents with the addition formula.
- practice Simplify $\sin^{-1}(\sin(2 \pi/3))$.
- practice Simplify $\cos^{-1}(\cos(4 \pi/3))$.
- practice Simplify $\tan^{-1}(\tan(3 \pi/4))$.
- practice Given $\sin^{-1}(x) = \pi/6$, use the complementary identity to find $\cos^{-1}(x)$.
- practice Given $\tan^{-1}(x) = \pi/6$, use the complementary identity to find $\cot^{-1}(x)$.
- practice Using $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$, evaluate $\cos^{-1}(-1/\sqrt{2})$.
- practice Using $\sin^{-1}(-x) = -\sin^{-1}(x)$, evaluate $\sin^{-1}(-\sqrt{3}/2)$.
- practice Using $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$, evaluate $\cot^{-1}(-1)$.
- practice Simplify $\cos^{-1}(\cos(7 \pi/6))$.
- practice Simplify $\sin^{-1}(\sin(7 \pi/6))$.
- practice Apply the addition formula to find $\tan^{-1}(1/5) + \tan^{-1}(2/3)$.
- practice Apply the addition formula to find $\tan^{-1}(2) + \tan^{-1}(-1/3)$.
- practice Which of these equals $\sin^{-1}(\sin(5 \pi/6))$?
- practice Simplify $\sin^{-1}(\sin(5 \pi/4))$.
- practice Evaluate $\tan^{-1}(1/6) + \tan^{-1}(5/7) + \cot^{-1}(1)$.
Answers
- $\pi/3$
- $2 \pi/3$
- $-\pi/4$
- $\pi/3$
- $\pi/3$
- $3 \pi/4$
- $-\pi/3$
- $3 \pi/4$
- $5 \pi/6$
- $-\pi/6$
- $\pi/4$
- $\pi/4$
- B — $\pi/6$.
- $-\pi/4$
- $\pi/2$
The Miscellaneous Exercise mixes every tool from the chapter into single multi-step problems. A typical problem reduces an out-of-branch angle, then applies a complementary or addition identity, then simplifies — without being told in advance which identity to reach for first.
- practice Evaluate $\sin^{-1}(\sin(5 \pi/3))$.
- practice Evaluate $\cos^{-1}(\cos(5 \pi/3))$.
- practice Evaluate $\tan^{-1}(\tan(2 \pi/3))$.
- practice If $\sin^{-1}(x) = \pi/5$, find $\cos^{-1}(x)$.
- practice If $\tan^{-1}(x) = 2 \pi/9$, find $\cot^{-1}(x)$.
- practice Evaluate $\sin^{-1}(-1/2) + \cos^{-1}(-1/2)$, and check it against the complementary identity.
- practice Apply the addition formula to find $\tan^{-1}(1/9) + \tan^{-1}(4/5)$.
- practice Apply the addition formula to find $\tan^{-1}(3) + \tan^{-1}(-1/2)$.
- practice Evaluate $\sec^{-1}(-2) + \csc^{-1}(\sqrt{2})$.
- practice Which of these equals $\cos^{-1}(-1) - \sin^{-1}(1)$?
- practice Simplify $\tan^{-1}(\tan(5 \pi/6))$.
- practice Evaluate $\sin^{-1}(\sin(5 \pi/6)) + \cos^{-1}(\cos(5 \pi/6))$.
- practice Evaluate $\sin^{-1}(\sin(11 \pi/6)) + \cos^{-1}(\cos(11 \pi/6))$.
- practice Evaluate $\tan^{-1}(1/9) + \tan^{-1}(4/5) + \cot^{-1}(1)$.
- practice Evaluate $\tan^{-1}(3) + \tan^{-1}(-1/2) + \sec^{-1}(-1)$.
Answers
- $-\pi/3$
- $\pi/3$
- $-\pi/3$
- $3 \pi/10$
- $5 \pi/18$
- $\pi/2$
- $\pi/4$
- $\pi/4$
- $11 \pi/12$
- B — $\pi/2$.
- $-\pi/6$
- $\pi$
- $0$
- $\pi/2$
- $5 \pi/4$