Basic concepts of inverse trigonometric functions

Why the ordinary trig functions cannot be inverted. A function $f$ has an inverse only if it is bijective — one-to-one (injective) and onto (surjective) — so that every output traces back to exactly one input. But $y = \sin x$ is periodic with period $2\pi$: $\sin 0 = \sin \pi = \sin 2\pi = 0$, so infinitely many $x$ give the same $y$. The same repetition happens for $\cos x, \tan x$ and the rest. Over $\mathbb{R}$ (or wherever each function is defined), none of the six trig functions is one-to-one, so none has a genuine inverse there.

The fix: restrict the domain. Pick a sub-interval of the natural domain on which the function is both one-to-one and covers its whole range exactly once. This chosen interval is the principal domain, and the corresponding restricted function is bijective — so it, and only it, gets inverted. For sine, the standard choice is $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$: on this interval $\sin x$ climbs steadily from $-1$ to $1$, hitting every value in $[-1,1]$ exactly once. That gives:

where: the domain $[-1,1]$ is exactly the range of $\sin x$ (sine never exceeds $1$ in size), and the range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ is exactly the principal domain chosen for sine. The same construction — choose a principal domain, invert only there — applies to all six functions:

Two conventions are worth locking in early. First, $\sin^{-1}x$ is functional notation for "the angle whose sine is $x$," never the reciprocal $\dfrac{1}{\sin x}$ — that reciprocal already has its own name, $\csc x$. Second, since $\sin^{-1}x$ is the inverse of (restricted) $\sin x$, its graph is the mirror image of $y = \sin x$ reflected across the line $y = x$; the same reflection relationship holds for every one of the six pairs.

There is no single formula to derive here — the domain/range pairs above are definitions, chosen so that inversion is possible at all — but the reasoning behind each choice is exactly what CBSE examines, so it earns the same careful, step-by-step treatment a derivation would get. This chain of reasoning applies to every one of the six functions, using $\tan x$ as the running example because it shows the full logic most cleanly, including a domain that excludes isolated points rather than just a periodic wraparound.

Step 1. Locate where the function is undefined or repeats. $\tan x = \dfrac{\sin x}{\cos x}$ is undefined wherever $\cos x = 0$, i.e. at $x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \dots$, and it is periodic with period $\pi$ (unlike sine and cosine, whose period is $2\pi$). Between consecutive undefined points, tangent runs through every real value exactly once.

Step 2. Choose one such interval as the principal domain. Take $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, the interval closest to the origin. On it, $\tan x$ increases continuously and without bound from $-\infty$ up to $+\infty$, so it is one-to-one (strictly increasing functions never repeat a value) and onto $\mathbb{R}$ (it reaches every real number).

Step 3. Read off the domain and range of the inverse from this bijection. Because $\tan x$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ hits every real number exactly once, its inverse $\tan^{-1}x$ is defined for every real input: domain $= \mathbb{R}$. And because the outputs of $\tan^{-1}x$ are exactly the inputs we started restricting, the range of $\tan^{-1}x$ is precisely the principal domain we chose: $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$.

Step 4. Same logic, different obstruction, for $\csc x$ and $\sec x$. Since $\csc x = 1/\sin x$ and $\sec x = 1/\cos x$, and $|\sin x| \le 1$, $|\cos x| \le 1$ always, it follows $|\csc x| \ge 1$ and $|\sec x| \ge 1$ wherever they are defined — neither ever lands strictly between $-1$ and $1$. That is why the domain of $\sec^{-1}x$ and $\csc^{-1}x$ is $\mathbb{R} - (-1, 1)$: it is not an arbitrary restriction but a direct consequence of what values secant and cosecant can actually take.

Verify, using the definitions above, that $\sec^{-1}(0.5)$ is undefined while $\sec^{-1}(2)$ is defined, and state the value's expected range.

Step 1. Recall the domain fact just derived: $\sec^{-1}x$ is defined only for $x \le -1$ or $x \ge 1$, i.e. $x \in \mathbb{R} - (-1, 1)$, because $|\sec x| \ge 1$ always.

Step 2. Test $x = 0.5$. Since $0.5$ lies strictly between $-1$ and $1$, it is in the excluded interval $(-1, 1)$, so $\sec^{-1}(0.5)$ does not exist — there is no angle whose secant equals $0.5$, because $\sec x$ never takes a value that small in magnitude.

Step 3. Test $x = 2$. Since $2 \ge 1$, it lies in the valid domain, so $\sec^{-1}(2)$ is defined. By the table, its value must lie in the range $[0, \pi] - \{\tfrac{\pi}{2}\}$; in fact $\sec\left(\tfrac{\pi}{3}\right) = 2$ and $\tfrac{\pi}{3} \in [0,\pi]-\{\tfrac{\pi}{2}\}$, consistent with the definition.

Final answer: $\sec^{-1}(0.5)$ is undefined; $\sec^{-1}(2)$ is defined and equals $\tfrac{\pi}{3}$, which lies in the expected range $[0,\pi]-\{\tfrac{\pi}{2}\}$.

Competitive stretch

Apply the same domain check to $\csc^{-1}(-1)$: since $-1 \le -1$, it lies in the valid domain $\mathbb{R} - (-1,1)$ (the endpoint $-1$ is included), so $\csc^{-1}(-1)$ is defined, and by the range table it must land in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right] - \{0\}$; indeed $\csc\left(-\tfrac{\pi}{2}\right) = -1$.

• Writing $\sin^{-1}x = \dfrac{1}{\sin x}$ — the $-1$ is functional inverse notation, never a reciprocal; the reciprocal of $\sin x$ is $\csc x$.
• Assuming $\sin^{-1}x$ or $\cos^{-1}x$ is defined for every real $x$ — both exist only for $x \in [-1, 1]$, since sine and cosine never exceed $1$ in absolute value.
• Forgetting that the inverse exists only after the domain of the original function is restricted; without that restriction the original function fails to be one-to-one and no inverse can be defined.
• Confusing domain with range: the domain of $\sin^{-1}x$ is an input restriction, $[-1,1]$; its range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ is an output restriction inherited from sine's principal domain — the two intervals are not interchangeable.

Ask the tutor: Want to see why the graph of $\cos^{-1}x$ is a decreasing curve while $\sin^{-1}x$ increases, or work through the domain-and-range logic for $\cot^{-1}x$ step by step?