📌 Snapshot
- Master principal value branches of sin⁻¹, cos⁻¹, tan⁻¹ and the rest, with worked examples and CUET traps.
- Why a trigonometric function has no inverse until its domain is restricted, and how each branch was chosen.
- Evaluating sin⁻¹(sin x) and cos⁻¹(cos x) when the angle falls outside the branch, the most repeated CUET pattern.
- The identities for 2 sin⁻¹ x, 3 sin⁻¹ x and cos⁻¹(−x), with the exact intervals where each one is valid.
📖 Detailed Notes
Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 2.
Which angle has sine one half?
$\sin 30^\circ = \dfrac{1}{2}$, but so is $\sin 150^\circ$, and $\sin 390^\circ$, and infinitely many more. If you ask "which angle has sine $\dfrac{1}{2}$?", there is no single answer, and a function must give exactly one output for each input. This chapter is the rule that forces a single answer: pick one fixed interval of angles, and always answer from inside it. Everything else, the six branches, the traps with angles outside the interval and the identities, follows from that one decision.
Why trigonometric functions need a restriction
A function has an inverse only if it is one-one (different inputs give different outputs) and onto (every value in the codomain is reached). Sine fails the first test badly: it repeats every $2\pi$, so a horizontal line at height $\dfrac{1}{2}$ cuts its graph infinitely often. The cure is to chop the domain down to a piece where the function runs through every value exactly once. That piece is called the principal value branch, and the angle the inverse returns from it is the principal value.
Whenever a question does not name a branch, the principal value branch is meant. The inverse functions are therefore single valued: $\sin^{-1}\dfrac{1}{2}$ is $\dfrac{\pi}{6}$ and nothing else, never "$\dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$".
Notation warning. $\sin^{-1} x$ means "the angle whose sine is $x$". It does not mean $(\sin x)^{-1}$, which is $\dfrac{1}{\sin x} = \operatorname{cosec} x$. The small $-1$ marks an inverse function, not a reciprocal.
The six branches in one table
Learn this table until you can rebuild it with the book shut. Almost every question in the chapter is a lookup in it.
| Inverse | Domain | Principal value branch | Trend |
|---|---|---|---|
| $\sin^{-1} x$ | $[-1, 1]$ | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ | Increasing |
| $\cos^{-1} x$ | $[-1, 1]$ | $[0, \pi]$ | Decreasing |
| $\tan^{-1} x$ | $\mathbb{R}$ | $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ | Increasing |
| $\cot^{-1} x$ | $\mathbb{R}$ | $(0, \pi)$ | Decreasing |
| $\sec^{-1} x$ | $\mathbb{R} - (-1, 1)$ | $[0, \pi] - \left\lbrace \dfrac{\pi}{2} \right\rbrace$ | Increasing on each piece |
| $\operatorname{cosec}^{-1} x$ | $\mathbb{R} - (-1, 1)$ | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \lbrace 0 \rbrace$ | Decreasing on each piece |
A way to remember it. The ranges come in two families. Sine, tangent and cosecant answer from the right half of the circle, so they can be negative and sit around zero. Cosine, cotangent and secant answer from the upper half, from $0$ to $\pi$, so they are never negative. A reciprocal pair always shares a family: sin with cosec, cos with sec, tan with cot. The excluded point is simply where the reciprocal function is undefined: $\operatorname{cosec}$ at $0$ and $\sec$ at $\dfrac{\pi}{2}$.
Why these branches and not others
Each branch has to meet three conditions: the function must be one-one on it, it must cover the whole range of values once, and it should contain the acute angles so that positive inputs give familiar answers.
| Function | Chosen interval | Why this one |
|---|---|---|
| $\sin$ | $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ | Runs from $-1$ up to $1$ without repeating. Contains $0$ and is symmetric about it |
| $\cos$ | $[0, \pi]$ | Runs from $1$ down to $-1$ once. The interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ will not work, because $\cos(-x) = \cos x$ makes it repeat |
| $\tan$ | $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ | Covers all of $\mathbb{R}$ once. The ends are open because tangent blows up there |
| $\cot$ | $(0, \pi)$ | Covers all of $\mathbb{R}$ once. Cotangent is undefined at $0$ and $\pi$ |
Other intervals would work equally well, for example $\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$ for sine. The ones above are a convention, which is exactly why you must memorise them rather than derive them.
The graph in one line. The graph of an inverse function is the mirror image of the original in the line $y = x$, because the roles of $x$ and $y$ swap. Domain and range swap too. Two consequences are worth keeping: $\tan^{-1} x$ flattens out towards the horizontal lines $y = \pm\dfrac{\pi}{2}$, and $\cot^{-1} x$ flattens out towards $y = 0$ and $y = \pi$.
Finding principal values
Read the table of standard angles backwards.
| Angle $\theta$ | $0$ | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ | $\dfrac{\pi}{3}$ | $\dfrac{\pi}{2}$ |
|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\dfrac{1}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{\sqrt{3}}{2}$ | $1$ |
| $\cos\theta$ | $1$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{1}{2}$ | $0$ |
| $\tan\theta$ | $0$ | $\dfrac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | undefined |
For a positive input, the reference angle in this table is already the answer. For a negative input, the two families behave differently, and this is the single most useful rule in the chapter.
| Inverse | Negative input rule | Example |
|---|---|---|
| $\sin^{-1}$, $\tan^{-1}$, $\operatorname{cosec}^{-1}$ | $f^{-1}(-x) = -f^{-1}(x)$ | $\sin^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$ |
| $\cos^{-1}$, $\cot^{-1}$, $\sec^{-1}$ | $f^{-1}(-x) = \pi - f^{-1}(x)$ | $\cos^{-1}\left(-\dfrac{1}{2}\right) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$ |
The right-half family goes negative; the upper-half family cannot, so it reflects the angle to $\pi - \theta$.
Why $\cos^{-1}(-x) = \pi - \cos^{-1} x$. Let $\theta = \cos^{-1} x$, so $\theta \in [0, \pi]$ and $\cos\theta = x$. Then $\cos(\pi - \theta) = -\cos\theta = -x$, and $\pi - \theta$ is also in $[0, \pi]$. So $\pi - \theta$ is the principal value of $\cos^{-1}(-x)$.
Worked example: adding two principal values. Find the value of $\sin^{-1}\left(-\dfrac{1}{2}\right) + \cos^{-1}\left(-\dfrac{1}{2}\right)$.
| Step | What we do | Result |
|---|---|---|
| 1 | Sine is in the right-half family: take the reference angle $\dfrac{\pi}{6}$ and make it negative | $\sin^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}$ |
| 2 | Cosine is in the upper-half family: reference angle $\dfrac{\pi}{3}$, reflect to $\pi - \dfrac{\pi}{3}$ | $\cos^{-1}\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$ |
| 3 | Add | $-\dfrac{\pi}{6} + \dfrac{4\pi}{6} = \dfrac{\pi}{2}$ |
Common wrong step. Writing $\cos^{-1}\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{3}$. A negative angle is outside $[0, \pi]$, so it can never be a cosine principal value.
Worked example: mixing two functions. Find $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$.
| Step | What we do | Result |
|---|---|---|
| 1 | $\tan\dfrac{\pi}{3} = \sqrt{3}$ and $\dfrac{\pi}{3}$ lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ | $\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$ |
| 2 | $\sec\theta = -2$ means $\cos\theta = -\dfrac{1}{2}$, and the angle must lie in $[0, \pi]$ | $\sec^{-1}(-2) = \dfrac{2\pi}{3}$ |
| 3 | Subtract | $\dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}$ |
Common wrong step. Reading $\sec^{-1}(-2)$ as $-\dfrac{\pi}{3}$ by treating secant like sine. Secant belongs to the upper-half family, so the negative input reflects to $\pi - \dfrac{\pi}{3}$.
Pause and predict: both $-\dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$ have tangent $-\sqrt{3}$. Which one is the principal value of $\tan^{-1}(-\sqrt{3})$?
Check your answer: $-\dfrac{\pi}{3}$. The tangent branch is the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$, and $\dfrac{2\pi}{3}$ lies outside it even though its tangent is correct.
The composite trap: $\sin^{-1}(\sin x)$ is not always $x$
Two statements look alike but behave very differently:
$$\sin(\sin^{-1} x) = x \quad \text{for every } x \in [-1, 1]$$
$$\sin^{-1}(\sin x) = x \quad \text{only for } x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$
The first is safe because $\sin^{-1} x$ always produces a genuine angle whose sine is $x$. The second fails because $\sin^{-1}$ can only answer from its branch, so if $x$ is outside the branch, it returns a different angle with the same sine. The same split holds for every pair: $\cos^{-1}(\cos x) = x$ only on $[0, \pi]$ and $\tan^{-1}(\tan x) = x$ only on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
The routine. To evaluate $\sin^{-1}(\sin\alpha)$, find the angle $\beta$ inside the branch that has the same sine as $\alpha$, using $\sin(\pi - \alpha) = \sin\alpha$ or a shift by $2\pi$. Do the same for the other functions with their own identities.
Worked example: three composites. Evaluate (a) $\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right)$, (b) $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$, (c) $\tan^{-1}\left(\tan\dfrac{2\pi}{3}\right)$.
| Part | What we do | Result |
|---|---|---|
| (a) | $\dfrac{3\pi}{5}$ is beyond $\dfrac{\pi}{2}$. Use $\sin\dfrac{3\pi}{5} = \sin\left(\pi - \dfrac{3\pi}{5}\right) = \sin\dfrac{2\pi}{5}$, and $\dfrac{2\pi}{5}$ is in the branch | $\dfrac{2\pi}{5}$ |
| (b) | $\dfrac{7\pi}{6}$ is beyond $\pi$. Use $\cos\dfrac{7\pi}{6} = \cos\left(2\pi - \dfrac{7\pi}{6}\right) = \cos\dfrac{5\pi}{6}$, and $\dfrac{5\pi}{6}$ is in $[0, \pi]$ | $\dfrac{5\pi}{6}$ |
| (c) | Tangent repeats every $\pi$: $\tan\dfrac{2\pi}{3} = \tan\left(\dfrac{2\pi}{3} - \pi\right) = \tan\left(-\dfrac{\pi}{3}\right)$, which is in the branch | $-\dfrac{\pi}{3}$ |
Common wrong step. Cancelling the function and its inverse to get the angle back: $\dfrac{3\pi}{5}$, $\dfrac{7\pi}{6}$, $\dfrac{2\pi}{3}$. Each of these is outside its own branch, so none of them can be an answer.
Pause and predict: is $\sin^{-1}(\sin 2)$ equal to $2$?
Check your answer: No. Here $2$ is in radians, about $114.6^\circ$, which is beyond $\dfrac{\pi}{2}$. Then $\sin 2 = \sin(\pi - 2)$ and $\pi - 2 \approx 1.14$ lies in the branch, so $\sin^{-1}(\sin 2) = \pi - 2$.
Beyond the textbook. For the exam: every answer must come from the branch in the table. In reality: the name "arc sine" is literal. On a circle of radius 1, an angle in radians equals the length of the arc it cuts off, so $\sin^{-1} x$ is "the arc whose sine is $x$". Calculators and programming languages (the asin, acos and atan functions) return exactly these principal values, which is why a calculator never gives you $150^\circ$ when you press inverse sine of $0.5$.
Domains of inverse functions
The input of an inverse trigonometric function must lie in its domain. For $\sin^{-1}$ and $\cos^{-1}$ that means the input has to lie in $[-1, 1]$.
Worked example: a domain. Find the domain of $f(x) = \sin^{-1}(2x - 1)$.
| Step | What we do | Result |
|---|---|---|
| 1 | The input $2x - 1$ must lie in $[-1, 1]$ | $-1 \le 2x - 1 \le 1$ |
| 2 | Add $1$ to every part | $0 \le 2x \le 2$ |
| 3 | Divide by $2$ | $0 \le x \le 1$, so the domain is $[0, 1]$ |
Common wrong step. Stopping at $-1 \le x \le 1$, as if the input were $x$ itself. The restriction applies to the whole expression inside the bracket.
Identities, and the intervals that make them true
Three standard simplifications are proved by substituting an angle and then checking that the angle lands in the right branch. The interval is the whole point of the result.
$$\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\sin^{-1} x \quad \text{for } -\dfrac{1}{\sqrt{2}} \le x \le \dfrac{1}{\sqrt{2}}$$
$$\sin^{-1}\left(2x\sqrt{1 - x^2}\right) = 2\cos^{-1} x \quad \text{for } \dfrac{1}{\sqrt{2}} \le x \le 1$$
Proof of the first case. Put $x = \sin\theta$ with $\theta = \sin^{-1} x$. Then $\sqrt{1 - x^2} = \cos\theta$, and $2x\sqrt{1 - x^2} = 2\sin\theta\cos\theta = \sin 2\theta$. Now $\sin^{-1}(\sin 2\theta) = 2\theta$ only if $2\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, that is $\theta \in \left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$, which is exactly $-\dfrac{1}{\sqrt{2}} \le x \le \dfrac{1}{\sqrt{2}}$. Second case: put $x = \cos\varphi$ with $\varphi \in \left[0, \dfrac{\pi}{4}\right]$ for $x \ge \dfrac{1}{\sqrt{2}}$. Then $2x\sqrt{1-x^2} = \sin 2\varphi$ with $2\varphi \in \left[0, \dfrac{\pi}{2}\right]$, giving $2\varphi = 2\cos^{-1} x$.
The triple-angle versions work the same way, from $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ and $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$:
$$3\sin^{-1} x = \sin^{-1}\left(3x - 4x^3\right) \quad \text{for } x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$$
$$3\cos^{-1} x = \cos^{-1}\left(4x^3 - 3x\right) \quad \text{for } x \in \left[\dfrac{1}{2}, 1\right]$$
The break points $\dfrac{1}{2}$ come from $3\theta$ having to stay inside $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ for sine (so $\theta \le \dfrac{\pi}{6}$), and inside $[0, \pi]$ for cosine (so $\theta \le \dfrac{\pi}{3}$).
Worked example: choosing the right form. Express $\sin^{-1}\dfrac{24}{25}$ in terms of $\sin^{-1}\dfrac{3}{5}$.
| Step | What we do | Result |
|---|---|---|
| 1 | Look for $x$ with $2x\sqrt{1 - x^2} = \dfrac{24}{25}$. Try $x = \dfrac{3}{5}$: then $\sqrt{1 - x^2} = \dfrac{4}{5}$ | $2 \times \dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$ |
| 2 | Is $x = \dfrac{3}{5} = 0.6$ inside $\left[-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right]$, where $\dfrac{1}{\sqrt{2}} \approx 0.707$? Yes | Use the first case |
| 3 | Apply it | $\sin^{-1}\dfrac{24}{25} = 2\sin^{-1}\dfrac{3}{5}$ |
Common wrong step. Choosing $x = \dfrac{4}{5}$, which also gives $2x\sqrt{1 - x^2} = \dfrac{24}{25}$, and writing $2\sin^{-1}\dfrac{4}{5}$. But $\dfrac{4}{5} > \dfrac{1}{\sqrt{2}}$, and $2\sin^{-1}\dfrac{4}{5} \approx 1.855$ is bigger than $\dfrac{\pi}{2}$, so it cannot be the output of $\sin^{-1}$. For $x = \dfrac{4}{5}$ the correct form is $2\cos^{-1}\dfrac{4}{5}$, which equals the same value as $2\sin^{-1}\dfrac{3}{5}$.
Beyond the textbook. For the exam: learn the interval with each identity, because options are built around the wrong interval. In reality: the same care is needed in calculus. When you differentiate $\sin^{-1}\left(2x\sqrt{1-x^2}\right)$, the answer is $\dfrac{2}{\sqrt{1-x^2}}$ on $\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right)$ but $-\dfrac{2}{\sqrt{1-x^2}}$ on $\left(\dfrac{1}{\sqrt{2}}, 1\right)$. The sign flips at the break point because the function switches from $2\sin^{-1}x$ to $2\cos^{-1}x$. Why it matters: the chapter on Continuity and Differentiability uses exactly these simplifications.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| $\sin^{-1}x = \dfrac{1}{\sin x}$ | The $-1$ looks like a power | It marks the inverse function. The reciprocal is $\operatorname{cosec}x$ |
| $\sin^{-1}(\sin x) = x$ for all $x$ | Function and inverse "cancel" | True only for $x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Reduce to the branch first |
| $\cot^{-1}\left(-\dfrac{1}{\sqrt{3}}\right) = -\dfrac{\pi}{3}$ | Negative input, so a negative angle | The branch is $(0, \pi)$, so the answer is $\dfrac{2\pi}{3}$ |
| $\cos^{-1}(-x) = -\cos^{-1}x$ | Sine and tangent behave this way | Cosine reflects: $\cos^{-1}(-x) = \pi - \cos^{-1}x$ |
| Range of $\sec^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \lbrace 0 \rbrace$ | It is the range of $\operatorname{cosec}^{-1}$ | $\sec^{-1}$ has $[0, \pi] - \left\lbrace\dfrac{\pi}{2}\right\rbrace$. Match the cos family to $[0, \pi]$ |
| $\tan^{-1}$ takes the value $\dfrac{\pi}{2}$ | The graph approaches it | The branch is open, so $\dfrac{\pi}{2}$ is never attained |
| $\sin^{-1}(2) = $ some angle | Looks like a harmless number | The domain is $[-1, 1]$, so $\sin^{-1}(2)$ is undefined for real numbers |
| $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ for all $x \in [-1, 1]$ | The algebra works for every $x$ | Valid only for $x \in \left[-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right]$ |
Formula sheet
| Result | Formula | Condition | When to use |
|---|---|---|---|
| Function after inverse | $\sin(\sin^{-1}x) = x$ | $x \in [-1, 1]$ | Always safe |
| Inverse after function | $\sin^{-1}(\sin x) = x$ | $x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ | Check the angle first |
| Cosine version | $\cos^{-1}(\cos x) = x$ | $x \in [0, \pi]$ | Reduce with $\cos(2\pi - x) = \cos x$ |
| Tangent version | $\tan^{-1}(\tan x) = x$ | $x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ | Shift by multiples of $\pi$ |
| Negative input, cos family | $\cos^{-1}(-x) = \pi - \cos^{-1}x$ | $x \in [-1, 1]$ | Principal values of negative inputs |
| Negative input, sin family | $\sin^{-1}(-x) = -\sin^{-1}x$ | $x \in [-1, 1]$ | Principal values of negative inputs |
| Double angle | $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x$ | $x \in \left[-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right]$ | Simplifying with $x = \sin\theta$ |
| Double angle, other end | $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\cos^{-1}x$ | $x \in \left[\dfrac{1}{\sqrt{2}}, 1\right]$ | Simplifying with $x = \cos\theta$ |
| Triple angle | $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ | $x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$ | Simplifying with $x = \sin\theta$ |
60-second revision
| Topic | Lock this in |
|---|---|
| Why restrict | Inverse needs one-one and onto, so choose one branch and answer only from it |
| Right-half family | $\sin^{-1}$: $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ · $\tan^{-1}$: open interval · $\operatorname{cosec}^{-1}$: same, minus $0$ |
| Upper-half family | $\cos^{-1}$: $[0, \pi]$ · $\cot^{-1}$: $(0, \pi)$ · $\sec^{-1}$: $[0, \pi]$ minus $\dfrac{\pi}{2}$ |
| Domains | $\sin^{-1}$, $\cos^{-1}$: $[-1, 1]$ · $\tan^{-1}$, $\cot^{-1}$: $\mathbb{R}$ · $\sec^{-1}$, $\operatorname{cosec}^{-1}$: $\mathbb{R} - (-1, 1)$ |
| Negative inputs | Sin, tan, cosec: put a minus sign · cos, cot, sec: $\pi$ minus the angle |
| Composite routine | Is the angle in the branch? If not, convert using $\pi - x$, $2\pi - x$ or a shift of $\pi$ |
| Trend | $\sin^{-1}$, $\tan^{-1}$ increase · $\cos^{-1}$, $\cot^{-1}$ decrease |
| Break points | $\dfrac{1}{\sqrt{2}}$ for the double angle · $\dfrac{1}{2}$ for the triple angle |
Where this chapter connects
- Previous: Relations and Functions supplies one-one, onto and invertible functions, which explain why branches are needed.
- Class XI link: Trigonometric Functions gives the standard angles, the double and triple angle formulas and the radian measure used throughout.
- Next: Matrices begins the algebra half of Class XII.
- Calculus links: Continuity and Differentiability differentiates all of these functions, and Integrals produces inverse trigonometric answers.
🎯 Practice MCQs
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Q1. The principal value of $\sin^{-1}\left(-\dfrac{1}{2}\right)$ is
▸ Show answer & explanation
Answer: B
The answer must lie in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Since $\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$, the value is $-\dfrac{\pi}{6}$. Option A has the wrong sign, and options C and D are outside the branch.
Q2. The principal value branch of $\operatorname{cosec}^{-1} x$ is
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Answer: C
Cosecant shares the family of sine, so its inverse answers from $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ with $0$ removed, since $\operatorname{cosec} 0$ is undefined. Option A is the range of $\sec^{-1}$, B is that of $\tan^{-1}$ and D is that of $\cot^{-1}$.
Q3. The value of $\cot^{-1}\left(-\dfrac{1}{\sqrt{3}}\right)$ is
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Answer: C
The range of $\cot^{-1}$ is $(0, \pi)$. Since $\cot^{-1}\dfrac{1}{\sqrt{3}} = \dfrac{\pi}{3}$, the negative input gives $\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$. Options A and D are negative, so they are outside the range, and B is the value for the positive input.
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Q4. The value of $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$ is
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Answer: D
$\tan^{-1}\sqrt{3} = \dfrac{\pi}{3}$ and $\sec^{-1}(-2) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$. The difference is $\dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}$.
Q5. The value of $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$ is
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Answer: C
The angle $\dfrac{7\pi}{6}$ is outside $[0, \pi]$. Using $\cos\dfrac{7\pi}{6} = \cos\left(2\pi - \dfrac{7\pi}{6}\right) = \cos\dfrac{5\pi}{6}$ and $\dfrac{5\pi}{6} \in [0, \pi]$, the value is $\dfrac{5\pi}{6}$.
Q6. The value of $\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right)$ is
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Answer: B
$\dfrac{3\pi}{5}$ is greater than $\dfrac{\pi}{2}$. Since $\sin\dfrac{3\pi}{5} = \sin\left(\pi - \dfrac{3\pi}{5}\right) = \sin\dfrac{2\pi}{5}$ and $\dfrac{2\pi}{5}$ lies in the branch, the value is $\dfrac{2\pi}{5}$. Option A forgets to reduce to the branch.
Q7. Consider the statements. (i) $\sin^{-1}x = \dfrac{1}{\sin x}$. (ii) $\sin(\sin^{-1}x) = x$ for every $x \in [-1, 1]$. (iii) $\sin^{-1}(\sin x) = x$ for every real $x$. (iv) The range of $\cos^{-1}$ is $[0, \pi]$. Which are correct?
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Answer: D
Statement (i) is false, since $\sin^{-1}x$ is the inverse and not the reciprocal. Statement (iii) is false, because it holds only for $x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Statements (ii) and (iv) are correct.
Q8. Match List I with List II. List I: (i) $\sin^{-1}$ (ii) $\cos^{-1}$ (iii) $\tan^{-1}$ (iv) $\cot^{-1}$. List II: (p) $(0, \pi)$ (q) $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (r) $[0, \pi]$ (s) $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
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Answer: D
Sine answers from the closed interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ and cosine from the closed $[0, \pi]$. Tangent uses the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ and cotangent the open $(0, \pi)$. So (i)-q, (ii)-r, (iii)-s, (iv)-p.
Q9. The domain of the function $\sin^{-1}(2x - 1)$ is
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Answer: D
The input must satisfy $-1 \le 2x - 1 \le 1$, which gives $0 \le 2x \le 2$ and hence $0 \le x \le 1$. Option A applies the condition to $x$ rather than to $2x - 1$.
Q10. If $0 < x < 1$ and $\cos^{-1}x = \alpha$, then $\cos^{-1}(-x)$ is equal to
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Answer: C
Since $\cos(\pi - \alpha) = -\cos\alpha = -x$ and $\pi - \alpha$ lies in $[0, \pi]$, the principal value of $\cos^{-1}(-x)$ is $\pi - \alpha$. Option A would be negative, which is impossible for $\cos^{-1}$.
Q11. The value of $\sin^{-1}\dfrac{24}{25}$ equals
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Answer: D
With $x = \dfrac{3}{5}$ we get $2x\sqrt{1 - x^2} = 2 \times \dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$, and $x = 0.6 \le \dfrac{1}{\sqrt{2}}$, so the value is $2\sin^{-1}\dfrac{3}{5}$. Options A and B are each about $1.855$, which is more than $\dfrac{\pi}{2}$, and an output of $\sin^{-1}$ cannot exceed $\dfrac{\pi}{2}$. Option C equals $\dfrac{\pi}{2}$ exactly, because the angles with sines $\dfrac{4}{5}$ and $\dfrac{3}{5}$ are complementary, while $\sin^{-1}\dfrac{24}{25}$ is about $1.287$.
Q12. Assertion (A): The principal value of $\tan^{-1}(-\sqrt{3})$ is $-\dfrac{\pi}{3}$ and not $\dfrac{2\pi}{3}$. Reason (R): The principal value branch of $\tan^{-1}$ is the open interval $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
▸ Show answer & explanation
Answer: C
Both $-\dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$ have tangent $-\sqrt{3}$, but only $-\dfrac{\pi}{3}$ lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. So R directly explains A.
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