NCERT Class 12 Mathematics — Inverse Trigonometric Functions, Exercise 2.2. All 15 questions solved.

Exercise 2.1 evaluated inverse trigonometric functions one at a time. This exercise composes them, and almost every question is solved by the same move: substitute a trigonometric function for the variable so the expression inside the inverse collapses to a single ratio of a single angle.

Which substitution to make is decided by what appears under the root:

What you see	Substitute
$\sqrt{1 + x^2}$	$x = \tan\theta$
$\sqrt{a^2 – x^2}$	$x = a\sin\theta$
$\sqrt{x^2 – a^2}$	$x = a\sec\theta$
$\dfrac{1 – x}{1 + x}$ or $\dfrac{1-\cos x}{1+\cos x}$	$x = \tan^2\theta$, or a half-angle identity

Alongside these, three identities do most of the work:

$$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1 – xy} \quad (xy < 1)$$

$$2\tan^{-1}x = \sin^{-1}\frac{2x}{1+x^2} = \cos^{-1}\frac{1-x^2}{1+x^2} = \tan^{-1}\frac{2x}{1-x^2}$$

Key insight. The domain restriction attached to each question is not decoration — it is what makes the answer a single clean expression instead of a case list. In question 5, the range $-\tfrac{\pi}{4} < x < \tfrac{3\pi}{4}$ is exactly the interval on which $\tfrac{\pi}{4} – x$ stays inside $\tan^{-1}$’s branch. Outside it, the answer would need $\pm\pi$ corrections. Read the restriction first; it usually tells you which form the answer takes.

Prove the following.

Question 1

$3\sin^{-1}x = \sin^{-1}\left(3x – 4x^3\right)$, $x \in \left[-\tfrac12, \tfrac12\right]$

Solution. Put $x = \sin\theta$, so $\theta = \sin^{-1}x$. The triple-angle identity gives

$$\sin 3\theta = 3\sin\theta – 4\sin^3\theta = 3x – 4x^3$$

Taking $\sin^{-1}$ of both sides gives $\sin^{-1}\left(3x – 4x^3\right) = 3\theta$ — but only if $3\theta$ lies in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$.

That is exactly what the restriction guarantees: $x \in \left[-\tfrac12, \tfrac12\right]$ means $\theta \in \left[-\tfrac{\pi}{6}, \tfrac{\pi}{6}\right]$, so $3\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$. Hence $3\theta = 3\sin^{-1}x$, as required.

Question 2

$3\cos^{-1}x = \cos^{-1}\left(4x^3 – 3x\right)$, $x \in \left[\tfrac12, 1\right]$

Solution. Put $x = \cos\theta$, so $\theta = \cos^{-1}x$, and use

$$\cos 3\theta = 4\cos^3\theta – 3\cos\theta = 4x^3 – 3x$$

For $\cos^{-1}(\cos 3\theta) = 3\theta$ we need $3\theta \in [0, \pi]$. With $x \in \left[\tfrac12, 1\right]$ we get $\theta \in \left[0, \tfrac{\pi}{3}\right]$ and so $3\theta \in [0, \pi]$, which is precisely the branch.

Write the following functions in the simplest form.

Question 3

$\tan^{-1}\dfrac{\sqrt{1+x^2} – 1}{x}$, $x \ne 0$

Solution. The $\sqrt{1+x^2}$ calls for $x = \tan\theta$, which makes $\sqrt{1+x^2} = \sec\theta$:

$$\frac{\sec\theta – 1}{\tan\theta} = \frac{1 – \cos\theta}{\sin\theta}$$

after multiplying numerator and denominator by $\cos\theta$. The half-angle forms $1 – \cos\theta = 2\sin^2\tfrac{\theta}{2}$ and $\sin\theta = 2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ reduce this to $\tan\tfrac{\theta}{2}$.

So the expression is $\tan^{-1}\left(\tan\tfrac{\theta}{2}\right) = \tfrac{\theta}{2}$.

Question 4

$\tan^{-1}\left(\sqrt{\dfrac{1 – \cos x}{1 + \cos x}}\right)$, $0 < x < \pi$

Solution. No substitution is needed — the half-angle identity is already staring at you:

$$\frac{1 – \cos x}{1 + \cos x} = \frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2}$$

The restriction $0 < x < \pi$ makes $\tfrac{x}{2} \in \left(0, \tfrac{\pi}{2}\right)$, where $\tan\tfrac{x}{2}$ is positive, so the square root is $+\tan\tfrac{x}{2}$ and no modulus is needed.

Question 5

$\tan^{-1}\left(\dfrac{\cos x – \sin x}{\cos x + \sin x}\right)$, $-\tfrac{\pi}{4} < x < \tfrac{3\pi}{4}$

Solution. Divide numerator and denominator by $\cos x$:

$$\frac{1 – \tan x}{1 + \tan x} = \frac{\tan\frac{\pi}{4} – \tan x}{1 + \tan\frac{\pi}{4}\tan x} = \tan\left(\frac{\pi}{4} – x\right)$$

Recognising $1$ as $\tan\tfrac{\pi}{4}$ is the whole trick. The given range of $x$ makes $\tfrac{\pi}{4} – x$ lie in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$, so the inverse cancels cleanly.

Question 6

$\tan^{-1}\dfrac{x}{\sqrt{a^2 – x^2}}$, $|x| < a$

Solution. The form $\sqrt{a^2 – x^2}$ calls for $x = a\sin\theta$, giving $\sqrt{a^2 – x^2} = a\cos\theta$:

$$\frac{a\sin\theta}{a\cos\theta} = \tan\theta$$

So the expression is $\theta = \sin^{-1}\dfrac{x}{a}$. The condition $|x| < a$ keeps $\theta$ inside $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ and the cosine positive.

Question 7

$\tan^{-1}\left(\dfrac{3a^2x – x^3}{a^3 – 3ax^2}\right)$, $a > 0$; $-\tfrac{a}{\sqrt3} < x < \tfrac{a}{\sqrt3}$

Solution. Divide numerator and denominator by $a^3$ to make the ratio $\tfrac{x}{a}$ appear:

$$\frac{3\left(\frac{x}{a}\right) – \left(\frac{x}{a}\right)^3}{1 – 3\left(\frac{x}{a}\right)^2}$$

With $\tfrac{x}{a} = \tan\theta$ this is exactly the triple-angle formula $\tan 3\theta = \dfrac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta}$.

The restriction $|x| < \tfrac{a}{\sqrt3}$ gives $|\tan\theta| < \tfrac{1}{\sqrt3}$, so $|\theta| < \tfrac{\pi}{6}$ and $3\theta$ stays in the branch.

Find the values of each of the following.

Question 8

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac12\right)\right]$

Solution. Work outwards from the innermost function. $\sin^{-1}\tfrac12 = \tfrac{\pi}{6}$, so $2\sin^{-1}\tfrac12 = \tfrac{\pi}{3}$ and $\cos\tfrac{\pi}{3} = \tfrac12$. The bracket is therefore $2 \times \tfrac12 = 1$, and $\tan^{-1}1 = \tfrac{\pi}{4}$.

Question 9

$\tan\dfrac12\left[\sin^{-1}\dfrac{2x}{1+x^2} + \cos^{-1}\dfrac{1-y^2}{1+y^2}\right]$, $|x| < 1$, $y > 0$ and $xy < 1$

Solution. Both bracketed terms are disguised doubles of an arctangent:

$$\sin^{-1}\frac{2x}{1+x^2} = 2\tan^{-1}x, \qquad \cos^{-1}\frac{1-y^2}{1+y^2} = 2\tan^{-1}y$$

The conditions $|x| < 1$ and $y > 0$ are exactly what makes each identity hold without a $\pi$ correction. So the bracket is $2\left(\tan^{-1}x + \tan^{-1}y\right)$ and the outer $\tfrac12$ cancels the $2$:

$$\tan\left(\tan^{-1}x + \tan^{-1}y\right) = \frac{x + y}{1 – xy}$$

with $xy < 1$ guaranteeing the addition formula applies directly.

Question 10

$\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right)$

Solution. The identity $\sin^{-1}(\sin\theta) = \theta$ holds only for $\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, and $\tfrac{2\pi}{3}$ is outside it. Replace the angle by one inside the branch with the same sine:

$$\sin\frac{2\pi}{3} = \sin\left(\pi – \frac{2\pi}{3}\right) = \sin\frac{\pi}{3}$$

and $\tfrac{\pi}{3}$ is in the branch.

Question 11

$\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$

Solution. $\tfrac{3\pi}{4}$ is outside $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Since tangent has period $\pi$:

$$\tan\frac{3\pi}{4} = \tan\left(\frac{3\pi}{4} – \pi\right) = \tan\left(-\frac{\pi}{4}\right)$$

and $-\tfrac{\pi}{4}$ is in the branch. Note that subtracting $\pi$ — not reflecting through $\pi$ as with sine — is what tangent’s periodicity allows.

Question 12

$\tan\left(\sin^{-1}\dfrac35 + \cot^{-1}\dfrac32\right)$

Solution. Convert both inverse functions to arctangents, since the outer function is a tangent. If $\alpha = \sin^{-1}\tfrac35$ then the opposite side is $3$ and the hypotenuse $5$, so the adjacent side is $4$ and $\tan\alpha = \tfrac34$. If $\beta = \cot^{-1}\tfrac32$ then $\tan\beta = \tfrac23$.

$$\tan(\alpha + \beta) = \frac{\frac34 + \frac23}{1 – \frac34 \cdot \frac23} = \frac{\frac{17}{12}}{\frac12}$$

Question 13

$\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$ is equal to

Solution. $\tfrac{7\pi}{6}$ is outside $[0, \pi]$, so option (A) is wrong however tempting. Cosine is even and has period $2\pi$:

$$\cos\frac{7\pi}{6} = \cos\left(2\pi – \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}$$

and $\tfrac{5\pi}{6} \in [0, \pi]$.

Question 14

$\sin\left(\dfrac{\pi}{3} – \sin^{-1}\left(-\dfrac12\right)\right)$ is equal to

Solution. $\sin^{-1}\left(-\tfrac12\right) = -\tfrac{\pi}{6}$, so the argument is

$$\frac{\pi}{3} – \left(-\frac{\pi}{6}\right) = \frac{\pi}{2}$$

and $\sin\tfrac{\pi}{2} = 1$. The double negative is the whole question — reading the inner value as $+\tfrac{\pi}{6}$ gives $\sin\tfrac{\pi}{6} = \tfrac12$, which is option (A).

Question 15

$\tan^{-1}\sqrt3 – \cot^{-1}\left(-\sqrt3\right)$ is equal to

Solution. $\tan^{-1}\sqrt3 = \tfrac{\pi}{3}$. For the second term, $\cot^{-1}$ has branch $(0, \pi)$, so a negative argument gives an obtuse angle:

$$\cot^{-1}\left(-\sqrt3\right) = \pi – \cot^{-1}\sqrt3 = \pi – \frac{\pi}{6} = \frac{5\pi}{6}$$

$$\frac{\pi}{3} – \frac{5\pi}{6} = -\frac{\pi}{2}$$

Taking $\cot^{-1}\left(-\sqrt3\right) = -\tfrac{\pi}{6}$ gives option (C), and is the mistake this question is built around.

A note on the numbering. On the printed page an instruction line reads “Find the values of each of the expressions in Exercises 16 to 18” immediately above question 10. That is a leftover from the pre-rationalisation edition, when this exercise ran to eighteen questions. It now has fifteen, and nothing is missing — questions 16 to 18 no longer exist.

Common mistakes

• Cancelling $\sin^{-1}(\sin\theta)$ to $\theta$ without checking the branch. Questions 10, 11 and 13 are all built on this. The identity holds only when $\theta$ is already inside the principal value branch.
• Using the wrong reduction to get back into the branch. Sine needs $\pi – \theta$, cosine needs $2\pi – \theta$, tangent needs $\theta – \pi$. Using sine’s rule on a cosine question gives option (D) in question 13.
• Ignoring the stated domain. In question 4 the restriction $0 < x < \pi$ is what allows $\sqrt{\tan^2\tfrac{x}{2}} = \tan\tfrac{x}{2}$ rather than $\left|\tan\tfrac{x}{2}\right|$.
• Applying $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\tfrac{x+y}{1-xy}$ when $xy > 1$. The formula then needs a $\pm\pi$. Question 9 states $xy < 1$ precisely so that it does not.
• Choosing the wrong substitution. $\sqrt{1+x^2}$ needs $x = \tan\theta$; using $x = \sin\theta$ leaves a root that will not simplify.
• Mishandling a negative argument in $\cot^{-1}$. As in Exercise 2.1, it returns an obtuse angle. Question 15 is the whole point.

Practise next

• Miscellaneous Exercise on Chapter 2 — the same identities used to prove results and solve equations.
• Exercise 2.1 — the principal value branches that every question here depends on.