Inverse Trigonometric Functions

Ask a calculator for $\sin 30°$ and it confidently returns $0.5$ . Ask it to run backward — for which angle is the sine equal to $0.5$ — and the honest answer is that infinitely many angles qualify: $30°$ , $150°$ , $390°$ , and so on forever in both directions. Sine, cosine and their relatives are periodic, so on their natural domains they are not one-one, and a function that is not one-one has no inverse. This chapter is the fix: by fencing each trigonometric function down to a carefully chosen interval on which it becomes one-one and onto, we recover a genuine inverse — and Class 12 asks you to master exactly where that fence sits for all six functions, because the fence, not the formula, is what the whole chapter tests. That fenced interval is called the principal value branch, and it is the single idea everything else in the chapter hangs off. $\sin^{-1} x$ is a function with domain $[-1, 1]$ and range $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ — no more, no less. Confuse the range with $[0, \pi]$ (the range that belongs to $\cos^{-1}$ instead) and an otherwise correct calculation quietly returns the wrong angle. Once the six principal branches are fixed, the chapter builds a toolkit of algebraic identities — complementary pairs like $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$ , sum formulas like $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ , and double-angle reductions — that let you simplify, evaluate and solve equations involving inverse trigonometric expressions without ever leaving algebra. Why does a chapter this short carry such weight in JEE and NEET? Because inverse trigonometric functions resurface constantly in calculus — as the antiderivatives of some of the most common rational and irrational integrands you will meet in Chapters 7 and 8 — and because their domain-restriction subtleties are a favourite trap for careless students. Master the six branches, the graph-by-reflection idea, and the identity toolkit here, and two entire future chapters get noticeably easier.

Key takeaways

An inverse trigonometric function exists only after restricting the domain of the original function to an interval where it is one-one and onto; that restricted range is called the principal value branch.
$\sin^{-1} : [-1,1] \to \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ , $\cos^{-1} : [-1,1] \to [0,\pi]$ , and $\tan^{-1} : \mathbb{R} \to \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ are the three branches used most often.
$\text{cosec}^{-1} : \mathbb{R}-(-1,1) \to \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]-\{0\}$ , $\sec^{-1} : \mathbb{R}-(-1,1) \to [0,\pi]-\{\tfrac{\pi}{2}\}$ , and $\cot^{-1} : \mathbb{R} \to (0,\pi)$ complete the set of six.
$\sin^{-1}x$ is never the same object as $(\sin x)^{-1} = \tfrac{1}{\sin x}$ ; the $-1$ superscript on an inverse trig symbol means "inverse function," not "reciprocal."
The graph of $y = f^{-1}(x)$ is always the mirror image of $y = f(x)$ in the line $y = x$ , obtained by swapping the roles of $x$ and $y$ .
$\sin(\sin^{-1}x) = x$ holds for every $x$ in $[-1,1]$ , but $\sin^{-1}(\sin x) = x$ holds only when $x$ already lies in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ — outside that interval you must first reduce $x$ into the branch.
Complementary identities pair the branches: $\sin^{-1}x+\cos^{-1}x = \tfrac{\pi}{2}$ , $\tan^{-1}x+\cot^{-1}x=\tfrac{\pi}{2}$ , and $\sec^{-1}x+\text{cosec}^{-1}x=\tfrac{\pi}{2}$ , each valid on the shared domain of both functions.
Negative-argument rules split into two families: $\sin^{-1}$ , $\tan^{-1}$ , $\text{cosec}^{-1}$ are odd ( $f^{-1}(-x)=-f^{-1}(x)$ ), while $\cos^{-1}$ , $\sec^{-1}$ , $\cot^{-1}$ satisfy $f^{-1}(-x)=\pi-f^{-1}(x)$ .
The addition formula $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ needs a correction of $+\pi$ or $-\pi$ whenever $xy>1$ , depending on the sign of $x$ and $y$ — the single most tested subtlety in the chapter.
Key idea: every inverse trigonometric function is the original circular function restricted to one principal branch on which it is bijective, so every simplification, identity and equation in this chapter is really a domain-and-range argument in algebraic disguise.

The core insight

Inverse trigonometric functions are ordinary sine, cosine and tangent restricted to principal value branches that make them bijective, and every identity in the chapter is valid only within — and because of — those exact domain and range restrictions.

What this chapter covers

The chapter builds in three clean stages, and it pays to keep the stages separate in your head. First comes the construction stage: starting from the six periodic trigonometric functions, restrict each domain to the smallest natural interval that makes it one-one and onto, and name the resulting inverse with its own domain and range — this is Section 2.2 of the NCERT text and gives you the six principal value branches you must memorise cold. Second is the graphing stage: because inverse functions are mirror images of the original across the line $y = x$ , you can sketch every inverse graph directly from the parent graph without fresh computation — this section also cements why the range had to be chosen the way it was. Third is the identity stage (Section 2.3): reciprocal relations connecting $\sin^{-1}$ with $\text{cosec}^{-1}$ , negative-argument rules, complementary-angle pairs summing to $\tfrac{\pi}{2}$ , and the addition/subtraction/double-angle formulas that let you collapse sums of inverse trig terms into a single angle — these are the working tools for both board simplification questions and JEE equation-solving. The worked examples below and the derivations that follow trace this same order: definitions and principal values first, then the two structural theorems (complementary relations and the addition formula) proved in full, then a double-angle reduction that recurs throughout the exercise set. Keep the domain of validity attached to every identity you write — an inverse trig identity without its domain note is, for examination purposes, an incomplete and often wrong statement.

The core ideas

Why a restriction is unavoidable

A function $f : X \to Y$ has an inverse $g : Y \to X$ only when $f$ is one-one (distinct inputs give distinct outputs) and onto (every element of $Y$ is hit). The sine function on all of $\mathbb{R}$ fails one-one badly — $\sin 0 = \sin \pi = \sin 2\pi = 0$ — so "the inverse of sine" does not exist as a single-valued function unless we first cut the domain down. Restrict $\sin$ to $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and it becomes strictly increasing from $-1$ to $1$ : one-one, and onto $[-1,1]$ . That restricted sine now has a genuine inverse, which we call $\sin^{-1}$ , read "inverse sine" or "arc sine," never "sine to the power minus one."

\sin^{-1} : [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. \qquad (1)

where: the domain $[-1,1]$ is exactly the range of ordinary sine (an inverse's domain is always its parent's range), and the range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ is called the principal value branch — the specific interval NCERT designates as "the" answer whenever a problem asks for $\sin^{-1}x$ without further qualification. Other intervals such as $\left[\tfrac{\pi}{2}, \tfrac{3\pi}{2}\right]$ also make sine bijective and would give a different, equally valid branch of the inverse, but examinations always default to the principal branch unless told otherwise. The same restriction argument runs for all six functions; only the surviving interval differs, chosen so it sits closest to the origin and avoids repeating a value. The table below is the anchor fact of the entire chapter.

Function	Domain	Principal value range
$\sin^{-1}x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1}x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1}x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1}x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1}x$	$\mathbb{R}-(-1,1)$	$[0,\pi]-\{\tfrac{\pi}{2}\}$
$\text{cosec}^{-1}x$	$\mathbb{R}-(-1,1)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]-\{0\}$

A note the NCERT text stresses on its own: $\sin^{-1}x$ should never be confused with $(\sin x)^{-1}$ . The latter simply means $\tfrac{1}{\sin x}$ , the reciprocal, and the two objects are unrelated except by a coincidence of notation — $\sin^{-1}(0.5) = \tfrac{\pi}{6}$ while $(\sin 0.5)^{-1} \approx 2.09$ , entirely different numbers of entirely different kinds.

Graphs as mirror images

If $(a, b)$ is a point on the graph of $y = \sin x$ restricted to $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ , then $(b, a)$ is the corresponding point on the graph of $y = \sin^{-1}x$ — swapping coordinates is exactly what "taking the inverse" means graphically. Equivalently, the graph of $y = \sin^{-1}x$ is the reflection of the restricted sine graph in the line $y = x$ . This single geometric fact lets you draw every inverse trig graph without a fresh table of values, once you already know the parent graph.

Figure

Graph of restricted sine and its inverse sine reflected across the line y equals x, showing the domain minus one to one mapping onto the range minus pi over two to pi over two

Reciprocal, negative-argument and complementary identities

Three families of identities connect the six inverse functions to each other, and every one of them is valid only for the domain stated alongside it — dropping the domain note is the single most common way marks are lost.

\sin^{-1}\frac{1}{x} = \text{cosec}^{-1}x, \quad \cos^{-1}\frac{1}{x} = \sec^{-1}x, \quad \tan^{-1}\frac{1}{x} = \cot^{-1}x \ \ (x > 0). \qquad (2)

where: these reciprocal relations hold for $|x| \ge 1$ in the first two, and for $x > 0$ in the third (for $x < 0$ , $\tan^{-1}\tfrac{1}{x} = \cot^{-1}x - \pi$ instead) — a direct consequence of matching principal ranges on both sides of each equation.

\sin^{-1}(-x) = -\sin^{-1}x, \quad \cos^{-1}(-x) = \pi - \cos^{-1}x, \quad \tan^{-1}(-x) = -\tan^{-1}x. \qquad (3)

where: $\sin^{-1}$ , $\tan^{-1}$ and $\text{cosec}^{-1}$ are odd functions because their principal ranges are symmetric about $0$ ; $\cos^{-1}$ , $\sec^{-1}$ and $\cot^{-1}$ instead satisfy the "supplementary" rule $f^{-1}(-x) = \pi - f^{-1}(x)$ because their ranges start at $0$ rather than being centred there.

\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad \sec^{-1}x + \text{cosec}^{-1}x = \frac{\pi}{2}. \qquad (4)

where: the first holds for every $x \in [-1,1]$ , the second for every $x \in \mathbb{R}$ , and the third for every $x$ with $|x| \ge 1$ — these are proved in full in the first derivation below and are the most frequently used identities in board simplification questions.

Sum, difference and double-angle formulas

For handling sums of two inverse tangents — extremely common in JEE algebra and in the calculus that follows this chapter — the key tool is

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right), \qquad xy < 1, \qquad (5)

where: the restriction $xy < 1$ keeps the sum of the two angles inside $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ so that the right-hand side's principal value matches the left-hand side exactly; when $xy > 1$ a correction term of $\pi$ or $-\pi$ must be added, proved case-by-case in the second derivation below. The matching subtraction formula is $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\tfrac{x-y}{1+xy}\right)$ for $xy > -1$ . Multiple-angle reductions turn a single inverse trig term with a doubled or tripled argument into a multiple of a simpler inverse trig term, and vice versa:

2\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} = \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{2x}{1-x^2}, \qquad |x| \le 1 \text{ (first form)}. \qquad (6)

where: each of the three right-hand expressions is exactly the double-angle formula for sine, cosine or tangent respectively, written with $x = \tan\theta$ ; the domain restriction on each differs slightly because the three parent functions have different principal ranges, which is precisely the subtlety explored in the third derivation below.

Figure

Graph of inverse tangent showing the S shaped curve rising from below to above the origin with horizontal asymptotes at y equals plus and minus pi over two

Derivations

Derivation AThe complementary relations between paired inverse functions

The goal here is to prove, from the definitions alone, the three identities $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$ for $x \in [-1,1]$ , $\tan^{-1}x + \cot^{-1}x = \tfrac{\pi}{2}$ for $x \in \mathbb{R}$ , and $\sec^{-1}x + \text{cosec}^{-1}x = \tfrac{\pi}{2}$ for $|x| \ge 1$ , which together form the most-used identity family in the chapter. The method is identical for all three: fix an angle equal to one of the inverse trig values, use the standard co-function identity relating sine and cosine of complementary angles, then argue that the resulting angle lands inside the correct principal branch so the inverse notation applies cleanly. We prove the sine-cosine case in full and note how the argument repeats for the other two pairs.

Step 1. Let $\sin^{-1}x = \theta$ for some $x \in [-1,1]$ . By definition of the principal branch, $\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and $\sin\theta = x$ .

Step 2. Use the co-function identity $\sin\theta = \cos\left(\tfrac{\pi}{2} - \theta\right)$ , which holds for every real $\theta$ as a direct consequence of the standard angle-subtraction formula for cosine. Substituting gives $\cos\left(\tfrac{\pi}{2} - \theta\right) = x$ .

Step 3. Check that the angle $\tfrac{\pi}{2} - \theta$ lies inside the principal branch of $\cos^{-1}$ , namely $[0,\pi]$ . Since $\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ , multiplying through by $-1$ and adding $\tfrac{\pi}{2}$ gives $\tfrac{\pi}{2} - \theta \in [0, \pi]$ exactly — the range check succeeds, so we may legitimately apply $\cos^{-1}$ to both sides of the equation from Step 2 and recover the angle itself rather than some other angle with the same cosine.

Step 4. Apply $\cos^{-1}$ to both sides: $\cos^{-1}(\cos(\tfrac{\pi}{2}-\theta)) = \cos^{-1}x$ . Because $\tfrac{\pi}{2}-\theta$ is already inside $[0,\pi]$ , the left side simplifies to $\tfrac{\pi}{2} - \theta$ itself, giving $\tfrac{\pi}{2} - \theta = \cos^{-1}x$ .

Step 5. Substitute back $\theta = \sin^{-1}x$ from Step 1: $\tfrac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$ , which rearranges to $\sin^{-1}x + \cos^{-1}x = \tfrac{\pi}{2}$ for every $x \in [-1,1]$ , as required.

Step 6. The other two pairs follow by an identical argument with the matching co-function identities: $\tan\theta = \cot\left(\tfrac{\pi}{2}-\theta\right)$ with $\theta \in \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ maps into $\left(0,\pi\right)$ , giving $\tan^{-1}x + \cot^{-1}x = \tfrac{\pi}{2}$ for all $x \in \mathbb{R}$ ; and $\sec\theta = \text{cosec}\left(\tfrac{\pi}{2}-\theta\right)$ with $\theta \in [0,\pi]-\{\tfrac{\pi}{2}\}$ maps into $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]-\{0\}$ , giving $\sec^{-1}x + \text{cosec}^{-1}x = \tfrac{\pi}{2}$ for all $|x| \ge 1$ . In every case the proof hinges on the same range-check in Step 3 — the identity only holds because the shifted angle lands exactly inside the partner function's principal branch.

Go deeper · Derivation B · 2 minThe addition formula for two inverse tangents, with its sign correction

Derivation BThe addition formula for two inverse tangents, with its sign correction

The goal is to derive $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ and to pin down exactly when a correction of $+\pi$ or $-\pi$ must be added, since blindly applying the bare formula is the single costliest error students make with this chapter in JEE-level algebra. The method starts from the ordinary tangent addition formula, which holds for underlying angles, then carefully checks which principal branch the sum of the two inverse-tangent angles actually falls into before naming that sum with inverse-tangent notation again.

Step 1. Let $\tan^{-1}x = \alpha$ and $\tan^{-1}y = \beta$ , so $\alpha, \beta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ , $\tan\alpha = x$ and $\tan\beta = y$ . We want a closed form for $\alpha + \beta$ .

Step 2. Apply the standard tangent addition formula $\tan(\alpha+\beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \dfrac{x+y}{1-xy}$ , valid whenever $1 - xy \neq 0$ , i.e. $xy \neq 1$ .

Step 3. Determine the range of $\alpha + \beta$ . Since each of $\alpha, \beta$ individually lies strictly between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2}$ , their sum lies strictly between $-\pi$ and $\pi$ — too wide an interval to be the principal branch of $\tan^{-1}$ directly, so we must split into cases based on where $\alpha+\beta$ actually falls.

Step 4. Case $xy < 1$ . Here it can be shown (by checking signs of $x$ and $y$ ) that $\alpha + \beta$ stays strictly inside $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ itself, so applying $\tan^{-1}$ to both sides of Step 2 simply returns $\alpha+\beta$ unchanged: $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ , exactly the formula quoted above, with no correction needed.

Step 5. Case $xy > 1$ with both $x > 0$ and $y > 0$ . Now both $\alpha$ and $\beta$ are individually closer to $\tfrac{\pi}{2}$ than to $0$ (a consequence of $x, y$ both being large enough that $xy$ exceeds $1$ ), so their sum $\alpha+\beta$ overshoots $\tfrac{\pi}{2}$ and lands in $\left(\tfrac{\pi}{2}, \pi\right)$ . Because $\tan^{-1}$ of the right-hand-side expression $\tfrac{x+y}{1-xy}$ (which is now negative, since $1-xy<0$ while $x+y>0$ ) can only return a value in $\left(-\tfrac{\pi}{2}, 0\right)$ , matching it to the true sum $\alpha+\beta \in \left(\tfrac{\pi}{2}, \pi\right)$ requires adding a full $\pi$ : $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ .

Step 6. Case $xy > 1$ with both $x < 0$ and $y < 0$ . By the mirror-image argument (both angles now close to $-\tfrac{\pi}{2}$ ), the sum lands in $\left(-\pi, -\tfrac{\pi}{2}\right)$ , and matching requires subtracting $\pi$ instead: $\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ . A quick numerical check with $x=y=-2$ confirms the pattern: $\tan^{-1}(-2)+\tan^{-1}(-2) \approx -2.214$ radians, while the bare formula $\tan^{-1}\!\left(\tfrac{-4}{1-4}\right)=\tan^{-1}(4/3)\approx 0.927$ radians is off by almost exactly $\pi$ , confirming the $-\pi$ correction is required here.

Go deeper · Derivation C · 1 minReducing a double inverse-sine angle to a single one

Derivation CReducing a double inverse-sine angle to a single one

The goal is to prove $\sin^{-1}\!\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x$ for $x$ restricted to $\left[-\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\right]$ , the method NCERT's own Example 3 uses and the template for every "express in simplest form" question built around a doubled or tripled inverse trig argument. The technique is always the same: substitute $x = \sin\theta$ so the messy algebraic expression collapses via a double-angle formula into a single trigonometric function of $2\theta$ , then check carefully that $2\theta$ lies inside the correct principal branch before removing the outer inverse-sine notation.

Step 1. Substitute $x = \sin\theta$ where $\theta = \sin^{-1}x \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ by definition. Then $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = |\cos\theta| = \cos\theta$ , since $\cos\theta \ge 0$ throughout $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ .

Step 2. Rewrite the argument of the outer inverse sine: $2x\sqrt{1-x^2} = 2\sin\theta\cos\theta$ . This is immediately recognisable as the right-hand side of the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$ , so the whole expression becomes $\sin^{-1}(\sin 2\theta)$ .

Step 3. Restrict $\theta$ further so the range check succeeds. We need $2\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ for $\sin^{-1}(\sin 2\theta)$ to simplify directly to $2\theta$ ; this forces $\theta \in \left[-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right]$ , which in terms of $x = \sin\theta$ means $x \in \left[-\tfrac{1}{\sqrt2}, \tfrac{1}{\sqrt2}\right]$ . Outside this narrower interval the identity still holds in spirit but the right-hand side must be replaced by $\pi - 2\sin^{-1}x$ or $-\pi-2\sin^{-1}x$ , exactly mirroring the sign-correction logic of the tangent addition formula above.

Step 4. With the range check satisfied, apply the simplification $\sin^{-1}(\sin 2\theta) = 2\theta$ directly, then substitute back $\theta = \sin^{-1}x$ : $\sin^{-1}\!\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x$ , valid for $-\tfrac{1}{\sqrt2} \le x \le \tfrac{1}{\sqrt2}$ , exactly as claimed.

Step 5. The companion identity follows by the same method with the substitution $x = \cos\theta$ instead: $\sin^{-1}\!\left(2x\sqrt{1-x^2}\right) = 2\cos^{-1}x$ for $\tfrac{1}{\sqrt2} \le x \le 1$ , since now $2x\sqrt{1-x^2} = 2\cos\theta\sin\theta = \sin 2\theta$ and the range check for $\cos^{-1}$ forces the complementary half of the domain. Together the two results cover the full domain $[-1,1]$ of the outer $\sin^{-1}$ , split by which of the two forms keeps the range check valid.

Worked examples

Board level - Reducing sin inverse of sin of an angle outside the principal branch

Step 1. Find the value of $\cos^{-1}\!\left(\cos\tfrac{13\pi}{6}\right)$ . The immediate temptation is to write the answer as $\tfrac{13\pi}{6}$ directly, but that is wrong because $\tfrac{13\pi}{6}$ does not lie in $[0,\pi]$ , the principal branch of $\cos^{-1}$ ; the identity $\cos^{-1}(\cos x) = x$ applies only when $x$ is already inside that branch.

Step 2. Reduce the angle. Since $\tfrac{13\pi}{6} = 2\pi + \tfrac{\pi}{6}$ , and cosine has period $2\pi$ , $\cos\tfrac{13\pi}{6} = \cos\tfrac{\pi}{6}$ .

Step 3. Check that $\tfrac{\pi}{6}$ lies inside $[0,\pi]$ — it does, comfortably. So $\cos^{-1}\!\left(\cos\tfrac{13\pi}{6}\right) = \cos^{-1}\!\left(\cos\tfrac{\pi}{6}\right) = \tfrac{\pi}{6}$ .

Step 4. State the answer. $\cos^{-1}\!\left(\cos\tfrac{13\pi}{6}\right) = \boldsymbol{\dfrac{\pi}{6}}$ .

Board level - Expressing an inverse tangent of a trigonometric ratio in simplest form

Step 1. Express $\tan^{-1}\!\left(\dfrac{\cos x}{1-\sin x}\right)$ in simplest form, for $-\tfrac{\pi}{2} < x < \tfrac{3\pi}{2}$ . Convert both numerator and denominator into half-angle form using $\cos x = \cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2}$ and $1 - \sin x = \left(\cos\tfrac{x}{2} - \sin\tfrac{x}{2}\right)^2$ (the latter from $1 = \cos^2\tfrac{x}{2}+\sin^2\tfrac{x}{2}$ and $\sin x = 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ ).

Step 2. Factor the numerator as a difference of squares: $\cos^2\tfrac{x}{2} - \sin^2\tfrac{x}{2} = \left(\cos\tfrac{x}{2}+\sin\tfrac{x}{2}\right)\left(\cos\tfrac{x}{2}-\sin\tfrac{x}{2}\right)$ . One factor of $\left(\cos\tfrac{x}{2}-\sin\tfrac{x}{2}\right)$ cancels against the matching factor in the denominator's square, leaving $\dfrac{\cos\tfrac{x}{2}+\sin\tfrac{x}{2}}{\cos\tfrac{x}{2}-\sin\tfrac{x}{2}}$ .

Step 3. Divide numerator and denominator by $\cos\tfrac{x}{2}$ to expose a tangent-addition pattern: $\dfrac{1+\tan\tfrac{x}{2}}{1-\tan\tfrac{x}{2}} = \tan\!\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)$ , using the standard addition identity $\tan(A+B) = \dfrac{\tan A+\tan B}{1-\tan A\tan B}$ with $A = \tfrac{\pi}{4}$ .

Step 4. Apply $\tan^{-1}$ and simplify, checking $\tfrac{\pi}{4}+\tfrac{x}{2}$ lies in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ over the stated domain: $\tan^{-1}\!\left(\dfrac{\cos x}{1-\sin x}\right) = \boldsymbol{\dfrac{\pi}{4}+\dfrac{x}{2}}$ .

Competitive level (JEE / NEET) - Solving a trigonometric equation involving two inverse functions

Step 1. Solve $2\tan^{-1}(\cos x) = \tan^{-1}(2\,\text{cosec}\,x)$ . Since the right side is a single inverse tangent, first collapse the left side into a single inverse tangent using the double-angle reduction $2\tan^{-1}t = \tan^{-1}\!\left(\dfrac{2t}{1-t^2}\right)$ with $t = \cos x$ , valid provided $|\cos x| < 1$ so that $2\tan^{-1}(\cos x)$ stays inside $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ .

Step 2. Write $2\tan^{-1}(\cos x) = \tan^{-1}\!\left(\dfrac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}\!\left(\dfrac{2\cos x}{\sin^2 x}\right)$ , using $1-\cos^2x=\sin^2x$ .

Step 3. Equate the two inverse-tangent arguments (legitimate once both sides are confirmed to lie in matching principal branches): $\dfrac{2\cos x}{\sin^2 x} = 2\,\text{cosec}\,x = \dfrac{2}{\sin x}$ .

Step 4. Cross-multiply: $2\cos x \sin x = 2\sin^2 x$ , i.e. $\sin x(\cos x - \sin x) = 0$ after dividing through by $2\sin x$ (valid since $\sin x \neq 0$ is required for $\text{cosec}\,x$ to be defined in the first place).

Step 5. Since $\sin x \neq 0$ is already required, the surviving factor gives $\cos x = \sin x$ , i.e. $\tan x = 1$ , so $x = \tfrac{\pi}{4}$ within $(0,\pi)$ where both $\cos x$ stays inside $(-1,1)$ and $\sin x > 0$ keeps $\text{cosec}\,x$ well defined with matching sign on both sides of the original equation.

Step 6. State the answer: $\boldsymbol{x = \dfrac{\pi}{4}}$ (and, by the periodicity of tangent together with the domain restrictions imposed by both inverse functions in the original equation, no other solution in a full period satisfies every restriction simultaneously). Timing note: this problem should take 3 to 4 minutes in a JEE setting — most of that time goes into correctly identifying and applying the double-angle collapse in Step 1, since attempting to solve the equation in its original two-inverse-function form directly is far slower and more error-prone.

Common traps

Board and competitive exams both mine the same handful of errors repeatedly, and every one of them traces back to a domain or range check that got skipped.

Treating $\sin^{-1}x$ as $\tfrac{1}{\sin x}$ . These are unrelated objects; the $-1$ on an inverse trig symbol denotes the inverse function, never a reciprocal power.
Applying $\sin^{-1}(\sin x) = x$ unconditionally. It only holds when $x$ already lies in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ ; outside that interval the angle must first be reduced into the principal branch, exactly as in the first worked example.
Memorising $\cos^{-1}$ 's range as $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ by confusing it with $\sin^{-1}$ . The correct range for $\cos^{-1}$ is $[0,\pi]$ ; mixing the two ranges is the single most common table-recall error.
Using the bare addition formula $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ whenever $xy > 1$ without the $\pm\pi$ correction, as shown in the second derivation — this alone accounts for a large share of lost marks in JEE algebra sections.
Forgetting the domain restriction on reciprocal identities. $\sin^{-1}\tfrac{1}{x}=\text{cosec}^{-1}x$ requires $|x|\ge 1$ ; plugging in $|x|<1$ produces a meaningless expression since $\tfrac1x$ would then exceed $1$ in magnitude on the left but the identity's derivation assumed $|x| \ge 1$ from the start.
Sign-flip mistakes in the negative-argument rules. $\sin^{-1}(-x)=-\sin^{-1}x$ but $\cos^{-1}(-x)=\pi-\cos^{-1}x$ — the second is not simply $-\cos^{-1}x$ , a slip that silently changes an answer's sign.
Ignoring the excluded point in $\sec^{-1}$ and $\text{cosec}^{-1}$ 's ranges. $\sec^{-1}$ never outputs $\tfrac{\pi}{2}$ and $\text{cosec}^{-1}$ never outputs $0$ , because those excluded values correspond exactly to $\cos x = 0$ or $\sin x = 0$ , where secant or cosecant themselves are undefined.

You should be able to…

State the domain and principal value range of all six inverse trigonometric functions without hesitation.
Explain, using one-one and onto, why an unrestricted trigonometric function cannot have an inverse.
Sketch the graph of any inverse trigonometric function as the reflection of its parent graph in the line $y=x$ .
Evaluate $\sin^{-1}(\sin x)$ , $\cos^{-1}(\cos x)$ and similar expressions correctly when $x$ lies outside the principal branch.
Apply the reciprocal, negative-argument and complementary identities with their correct domains attached.
Use the addition and subtraction formulas for $\tan^{-1}x \pm \tan^{-1}y$ , including the sign correction when $xy>1$ .
Reduce double- and triple-angle inverse trig expressions to a single inverse trig term using trigonometric substitution.
Solve simple equations that mix two or more inverse trigonometric functions, checking the domain of every intermediate step.
Distinguish $\sin^{-1}x$ from $(\sin x)^{-1}$ and $\text{cosec}\,x$ in every context without conflating them.

Ask the tutor: Want to work through a few more xy-greater-than-one addition-formula equations until the sign correction becomes automatic?

Why does cos inverse x have range 0 to pi while sin inverse x has range minus pi over 2 to pi over 2

The choice always aims for the interval closest to zero on which the function is one-one and onto. Cosine restricted to $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ is not one-one, because $\cos(-\theta)=\cos\theta$ for every $\theta$ in that interval — it fails the horizontal line test immediately. The nearest interval to zero on which cosine is genuinely one-one and covers all of $[-1,1]$ is $[0,\pi]$ , so that becomes the principal branch by construction, not by arbitrary convention.

Is sin inverse x the same thing as cosec x

No, and this confusion costs marks every year. $\sin^{-1}x$ is the inverse function of sine, an angle whose sine equals $x$ . $\text{cosec}\,x$ is a completely different trigonometric ratio, equal to $\tfrac{1}{\sin x}$ . They are connected only through the separate identity $\sin^{-1}\tfrac1x = \text{cosec}^{-1}x$ for $|x|\ge 1$ , which relates two different inverse functions to each other — it does not make $\sin^{-1}$ and $\text{cosec}$ interchangeable in any other context.

Why do some tan inverse sum formulas need an extra plus or minus pi added

Because $\tan^{-1}x$ and $\tan^{-1}y$ are each confined to $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ , their sum can range anywhere across almost the full interval $(-\pi,\pi)$ , which is wider than the principal branch of $\tan^{-1}$ itself. When $xy>1$ the true sum has slipped outside $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ , but the plain formula $\tan^{-1}\!\left(\tfrac{x+y}{1-xy}\right)$ can only ever output a value inside that narrower branch, so a correction of $+\pi$ or $-\pi$ is needed to land back on the true sum. The full case analysis is worked out in the second derivation in this article.

How should I memorise all six domain and range pairs before an exam

Learn them in three complementary pairs rather than six separate facts: sin inverse and cosec inverse share the shape $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ , with cosec inverse excluding zero because cosecant is undefined there; cos inverse and sec inverse share the shape $[0,\pi]$ , with sec inverse excluding $\tfrac{\pi}{2}$ for the same reason; and tan inverse and cot inverse share open intervals, $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $(0,\pi)$ respectively, both open because tangent and cotangent themselves blow up at the endpoints. Three shapes, not six, is far easier to hold under exam pressure.

Do I need calculus for this chapter

No — Chapter 2 itself is entirely algebraic: definitions, graphs by reflection, and identity manipulation using ordinary trigonometric formulas you already know from Class 11. Calculus enters only later, in the differentiation and integration chapters, where the derivatives of inverse trigonometric functions (for instance the derivative of $\sin^{-1}x$ is $\tfrac{1}{\sqrt{1-x^2}}$ ) and their use as standard integral forms become essential tools — this chapter is the algebraic foundation those later chapters lean on.