NCERT Class 12 Mathematics — Inverse Trigonometric Functions, Exercise 2.1. All 14 questions solved.

Every question in this exercise is answered by the same two-line method: find an angle whose trigonometric ratio is the given number, then check that the angle lies in the principal value branch. If it does not, replace it by one that does.

The branches are worth memorising before starting, because half the marks in this exercise are lost by using the wrong one:

Key insight. The branches split into two families. $\sin^{-1}$, $\operatorname{cosec}^{-1}$ and $\tan^{-1}$ live in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and so return negative angles for negative arguments. $\cos^{-1}$, $\sec^{-1}$ and $\cot^{-1}$ live in $[0, \pi]$ and so return obtuse angles for negative arguments — never negative ones.

That single distinction decides questions 4, 5, 6, 9 and 10, and it is the reason $\cos^{-1}\left(-\tfrac12\right)$ is $\tfrac{2\pi}{3}$ and not $-\tfrac{\pi}{3}$.

Find the principal values of the following.

Question 1

$\sin^{-1}\left(-\dfrac{1}{2}\right)$

Solution. Let $\sin^{-1}\left(-\tfrac12\right) = y$, so $\sin y = -\tfrac12$. The branch is $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, and within it

$$\sin\left(-\frac{\pi}{6}\right) = -\frac12$$

Question 2

$\cos^{-1}\left(\dfrac{\sqrt3}{2}\right)$

Solution. $\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$, and $\tfrac{\pi}{6}$ lies in $[0, \pi]$.

Question 3

$\operatorname{cosec}^{-1}(2)$

Solution. $\operatorname{cosec} y = 2$ means $\sin y = \tfrac12$, so $y = \tfrac{\pi}{6}$, which lies in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] – \{0\}$.

Converting to sine first is always the safer route — the cosecant graph is harder to picture than the sine graph.

Question 4

$\tan^{-1}\left(-\sqrt3\right)$

Solution. $\tan\tfrac{\pi}{3} = \sqrt3$, and tangent is an odd function, so $\tan\left(-\tfrac{\pi}{3}\right) = -\sqrt3$. Since $-\tfrac{\pi}{3}$ lies in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, it is the principal value.

Question 5

$\cos^{-1}\left(-\dfrac{1}{2}\right)$

Solution. Here the sign changes the method. $\cos^{-1}$ never returns a negative angle, so instead of reflecting through $0$ we reflect through $\tfrac{\pi}{2}$ using $\cos(\pi – \theta) = -\cos\theta$:

$$\cos\left(\pi – \frac{\pi}{3}\right) = -\frac12 \quad\Longrightarrow\quad y = \frac{2\pi}{3}$$

and $\tfrac{2\pi}{3} \in [0, \pi]$.

Question 6

$\tan^{-1}(-1)$

Solution. $\tan\left(-\tfrac{\pi}{4}\right) = -1$, and $-\tfrac{\pi}{4}$ is in the branch.

Question 7

$\sec^{-1}\left(\dfrac{2}{\sqrt3}\right)$

Solution. $\sec y = \tfrac{2}{\sqrt3}$ means $\cos y = \tfrac{\sqrt3}{2}$, so $y = \tfrac{\pi}{6}$, which lies in $[0, \pi] – \left\{\tfrac{\pi}{2}\right\}$.

Question 8

$\cot^{-1}\left(\sqrt3\right)$

Solution. $\cot y = \sqrt3$ means $\tan y = \tfrac{1}{\sqrt3}$, so $y = \tfrac{\pi}{6} \in (0, \pi)$.

Question 9

$\cos^{-1}\left(-\dfrac{1}{\sqrt2}\right)$

Solution. As in question 5, use $\cos(\pi – \theta) = -\cos\theta$ with $\theta = \tfrac{\pi}{4}$:

$$\cos\left(\pi – \frac{\pi}{4}\right) = \cos\frac{3\pi}{4} = -\frac{1}{\sqrt2}$$

Question 10

$\operatorname{cosec}^{-1}\left(-\sqrt2\right)$

Solution. $\operatorname{cosec} y = -\sqrt2$ means $\sin y = -\tfrac{1}{\sqrt2}$. Cosecant belongs to the $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ family, so the answer is negative:

$$\sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt2}$$

Compare with question 9: the same magnitude of argument, opposite family, and therefore a completely different answer.

Find the values of the following.

Question 11

$\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right)$

Solution. Evaluate each term in its own branch, then add:

$$\tan^{-1}(1) = \frac{\pi}{4}, \qquad \cos^{-1}\left(-\frac12\right) = \frac{2\pi}{3}, \qquad \sin^{-1}\left(-\frac12\right) = -\frac{\pi}{6}$$

The last two combine to $\tfrac{2\pi}{3} – \tfrac{\pi}{6} = \tfrac{\pi}{2}$.

Question 12

$\cos^{-1}\dfrac{1}{2} + 2\sin^{-1}\dfrac{1}{2}$

Solution. $\cos^{-1}\tfrac12 = \tfrac{\pi}{3}$ and $\sin^{-1}\tfrac12 = \tfrac{\pi}{6}$, so the expression is

$$\frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3}$$

Question 13

If $\sin^{-1}x = y$, then

Solution. This is the definition of the principal value branch of $\sin^{-1}$, and the question turns on whether the endpoints are included. They are: $\sin^{-1}(1) = \tfrac{\pi}{2}$ and $\sin^{-1}(-1) = -\tfrac{\pi}{2}$ are both defined, so the interval is closed.

Option (D) is the branch of $\tan^{-1}$, which is open because $\tan\left(\pm\tfrac{\pi}{2}\right)$ does not exist. Option (A) is the branch of $\cos^{-1}$.

Question 14

$\tan^{-1}\sqrt3 – \sec^{-1}(-2)$ is equal to

Solution. $\tan^{-1}\sqrt3 = \tfrac{\pi}{3}$. For the second term, $\sec y = -2$ means $\cos y = -\tfrac12$, and $\sec^{-1}$ belongs to the $[0, \pi]$ family, so $y = \tfrac{2\pi}{3}$ — not $-\tfrac{\pi}{3}$.

$$\frac{\pi}{3} – \frac{2\pi}{3} = -\frac{\pi}{3}$$

Option (D) is what you get by taking $\sec^{-1}(-2) = -\tfrac{\pi}{3}$, the error this exercise exists to prevent.

Common mistakes

• Giving a negative angle for $\cos^{-1}$, $\sec^{-1}$ or $\cot^{-1}$ of a negative number. These three never return a negative value. Their answers to negative arguments are obtuse: between $\tfrac{\pi}{2}$ and $\pi$.
• Using $\cos(-\theta) = \cos\theta$ to handle a negative argument. That identity is true but useless here, because it produces an angle outside the branch. The identity you need is $\cos(\pi – \theta) = -\cos\theta$.
• Confusing $\sin^{-1}x$ with $(\sin x)^{-1}$. The first is an angle; the second is $\tfrac{1}{\sin x}$. NCERT flags this in a note immediately before the exercise for a reason.
• Answering with any correct angle rather than the principal one. $\sin\tfrac{5\pi}{6} = \tfrac12$ is true, but $\sin^{-1}\tfrac12$ is $\tfrac{\pi}{6}$, because that is the value in the branch.
• Forgetting that $\tan^{-1}$’s branch is open. Question 13 hinges on $\sin^{-1}$’s being closed while $\tan^{-1}$’s is not.

Practise next

• Exercise 2.2 — the identities that let you simplify a composite inverse trigonometric expression before evaluating it.
• Miscellaneous Exercise on Chapter 2 — expressions like $\cos^{-1}\left(\cos\tfrac{13\pi}{6}\right)$, where the inner angle is outside the branch.