NCERT Exemplar Solutions Class 12 Maths Inverse Trigonometric Functions — Objective Type Q20–37

NCERT Exemplar Problems, Class 12 Mathematics — Inverse Trigonometric Functions, Objective Type questions 20 to 37. All 18 questions solved.

Eighteen multiple-choice questions. The first group tests the principal value branches themselves (20, 21) and domains (24, 25); the middle group asks you to evaluate an expression like $\sin^{-1}\left(\cos\frac{33\pi}{5}\right)$ by moving the angle into a branch (23, 28, 29, 32, 33); and the last group uses the identities $\sin^{-1}x + \cos^{-1}x = \frac\pi2$ and $\tan^{-1}x + \cot^{-1}x = \frac\pi2$ to turn an equation in two inverse functions into one (22, 26, 30, 37).

One answer in the printed key is wrong, Question 37. It is explained where it occurs.

Key insight. $\sin^{-1}x + \cos^{-1}x = \frac\pi2$ and $\tan^{-1}x + \cot^{-1}x = \frac\pi2$ for every $x$ in the domain. Whenever a question mixes a function with its co-function, replace one by $\frac\pi2$ minus the other first; the equation then has only one unknown function in it.

Question 20

Which of the following is the principal value branch of $\cos^{-1}x$?

Solution. $\cos$ is one-one on $[0, \pi]$ and takes every value in $[-1, 1]$ there, from $\cos0 = 1$ down to $\cos\pi = -1$. Both endpoints are needed: $\cos^{-1}1 = 0$ and $\cos^{-1}(-1) = \pi$. Option (A) is the branch of $\sin^{-1}$; option (B) is the branch of $\cot^{-1}$.

Question 21

Which of the following is the principal value branch of $\mathrm{cosec}^{-1}x$?

Solution. $\mathrm{cosec}\,y = \frac{1}{\sin y}$, so it uses the same interval as $\sin^{-1}$, $\left[-\frac\pi2, \frac\pi2\right]$, but $y = 0$ has to go because $\sin0 = 0$ and $\mathrm{cosec}\,0$ is undefined. Option (B) is the branch of $\sec^{-1}$, which removes $\frac\pi2$ for the same reason.

Question 22

If $3\tan^{-1}x + \cot^{-1}x = \pi$, then $x$ equals

Solution. Replace $\cot^{-1}x$ by $\frac\pi2 – \tan^{-1}x$:

$$3\tan^{-1}x + \frac\pi2 – \tan^{-1}x = \pi \implies 2\tan^{-1}x = \frac\pi2 \implies \tan^{-1}x = \frac\pi4 \implies x = 1.$$

Question 23

The value of $\sin^{-1}\left(\cos\left(\dfrac{33\pi}{5}\right)\right)$ is

Solution. Remove whole periods first: $\frac{33\pi}{5} = 6\pi + \frac{3\pi}{5}$, so $\cos\frac{33\pi}{5} = \cos\frac{3\pi}{5}$. Then use $\sin^{-1}y = \frac\pi2 – \cos^{-1}y$, and $\cos^{-1}\left(\cos\frac{3\pi}{5}\right) = \frac{3\pi}{5}$ because $\frac{3\pi}{5} \in [0, \pi]$:

$$\sin^{-1}\left(\cos\frac{3\pi}{5}\right) = \frac\pi2 – \frac{3\pi}{5} = -\frac{\pi}{10}.$$

As a check, $\cos\frac{3\pi}{5}$ is negative (the angle is past $\frac\pi2$), so the answer must be negative. Options (A) and (B) are not even in $\left[-\frac\pi2, \frac\pi2\right]$.

Question 24

The domain of the function $\cos^{-1}(2x – 1)$ is

Solution. $\cos^{-1}$ accepts inputs in $[-1, 1]$, so we need $-1 \le 2x – 1 \le 1$, i.e. $0 \le 2x \le 2$, i.e. $0 \le x \le 1$. Option (D) is the range of $\cos^{-1}$, not a domain.

Question 25

The domain of the function defined by $f(x) = \sin^{-1}\sqrt{x – 1}$ is

Solution. Two conditions. The square root needs $x – 1 \ge 0$, so $x \ge 1$. Then $\sin^{-1}$ needs $\sqrt{x – 1} \le 1$ (it is automatically $\ge 0 > -1$), so $x – 1 \le 1$ and $x \le 2$.

Question 26

If $\cos\left(\sin^{-1}\dfrac25 + \cos^{-1}x\right) = 0$, then $x$ is equal to

Solution. $\sin^{-1}\frac25$ is between $0$ and $\frac\pi2$ and $\cos^{-1}x$ is between $0$ and $\pi$, so the sum lies in $\left(0, \frac{3\pi}{2}\right)$. The only angle there with cosine $0$ is $\frac\pi2$. So $\cos^{-1}x = \frac\pi2 – \sin^{-1}\frac25$, and by the identity $\sin^{-1}y + \cos^{-1}y = \frac\pi2$ that is $\cos^{-1}\frac25$. Hence $x = \frac25$.

Question 27

The value of $\sin\left(2\tan^{-1}(0.75)\right)$ is equal to

Solution. With $\tan\theta = \frac34$, $\sin2\theta = \dfrac{2\tan\theta}{1 + \tan^2\theta} = \dfrac{\frac32}{\frac{25}{16}} = \dfrac{24}{25} = 0.96$. Option (D) is the slip of doubling inside: $2\tan^{-1}0.75$ is not $\tan^{-1}1.5$, and certainly not $1.5$.

Question 28

The value of $\cos^{-1}\left(\cos\dfrac{3\pi}{2}\right)$ is equal to

Solution. $\cos\frac{3\pi}{2} = 0$, and $\cos^{-1}0 = \frac\pi2$. Option (B) is the early cancellation: $\frac{3\pi}{2}$ is outside $[0, \pi]$, so $\cos^{-1}$ cannot return it. The other two options are also outside the range.

Question 29

The value of the expression $2\sec^{-1}2 + \sin^{-1}\left(\dfrac12\right)$ is

Solution. $\sec^{-1}2$ is the angle in $[0, \pi]$ with cosine $\frac12$, namely $\frac\pi3$. $\sin^{-1}\frac12 = \frac\pi6$. So the expression is $\frac{2\pi}{3} + \frac\pi6 = \frac{5\pi}{6}$.

Question 30

If $\tan^{-1}x + \tan^{-1}y = \dfrac{4\pi}{5}$, then $\cot^{-1}x + \cot^{-1}y$ equals

Solution. Use $\cot^{-1}t = \frac\pi2 – \tan^{-1}t$ for each:

$$\cot^{-1}x + \cot^{-1}y = \pi – \left(\tan^{-1}x + \tan^{-1}y\right) = \pi – \frac{4\pi}{5} = \frac\pi5.$$

Question 31

If $\sin^{-1}\left(\dfrac{2a}{1 + a^2}\right) + \cos^{-1}\left(\dfrac{1 – a^2}{1 + a^2}\right) = \tan^{-1}\left(\dfrac{2x}{1 – x^2}\right)$, where $a, x \in\, ]0, 1[$, then the value of $x$ is

Solution. Put $a = \tan\theta$ with $\theta \in \left(0, \frac\pi4\right)$. Then $\dfrac{2a}{1 + a^2} = \sin2\theta$ and $\dfrac{1 – a^2}{1 + a^2} = \cos2\theta$, with $2\theta \in \left(0, \frac\pi2\right)$ inside both branches, so the left side is $2\theta + 2\theta = 4\tan^{-1}a$. Likewise, for $x \in (0, 1)$ the right side is $2\tan^{-1}x$. So

$$2\tan^{-1}x = 4\tan^{-1}a \implies \tan^{-1}x = 2\tan^{-1}a \implies x = \tan(2\tan^{-1}a) = \frac{2a}{1 – a^2}.$$

Option (C) comes from treating the two left-hand terms as if they added to $2\tan^{-1}a$ rather than $4\tan^{-1}a$. (For $x$ to stay inside $]0, 1[$ as the question requires, $a$ has to be below $\sqrt2 – 1$; the question leaves that unsaid.)

Question 32

The value of $\cot\left[\cos^{-1}\left(\dfrac{7}{25}\right)\right]$ is

Solution. $\cos^{-1}\frac{7}{25}$ is an acute angle with adjacent side 7 and hypotenuse 25, so the opposite side is $\sqrt{625 – 49} = 24$. Its cotangent is adjacent over opposite, $\frac{7}{24}$.

Question 33

The value of the expression $\tan\left(\dfrac12\cos^{-1}\dfrac{2}{\sqrt5}\right)$ is

[Hint: $\tan\dfrac\theta2 = \sqrt{\dfrac{1 – \cos\theta}{1 + \cos\theta}}$]

Solution. With $\cos\theta = \frac{2}{\sqrt5}$ and $\theta$ acute, $\frac\theta2$ is acute and the positive root is the right one:

$$\tan\frac\theta2 = \sqrt{\frac{1 – \frac{2}{\sqrt5}}{1 + \frac{2}{\sqrt5}}} = \sqrt{\frac{\sqrt5 – 2}{\sqrt5 + 2}} = \sqrt{\frac{(\sqrt5 – 2)^2}{5 – 4}} = \sqrt5 – 2.$$

Option (A) is the reciprocal, $\cot\frac\theta2$. A quick check: $\sqrt5 – 2 \approx 0.24$, which is less than $1$, as the tangent of an angle under $\frac\pi4$ must be.

Question 34

If $|x| \le 1$, then $2\tan^{-1}x + \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$ is equal to

Solution. Put $x = \tan\theta$; $|x| \le 1$ means $\theta \in \left[-\frac\pi4, \frac\pi4\right]$, so $2\theta \in \left[-\frac\pi2, \frac\pi2\right]$, inside the branch of $\sin^{-1}$. Then $\sin^{-1}\dfrac{2x}{1 + x^2} = \sin^{-1}(\sin2\theta) = 2\theta = 2\tan^{-1}x$, and the whole expression is $4\tan^{-1}x$.

The condition $|x| \le 1$ matters. For $x > 1$ the same expression is the constant $\pi$, which is why option (D) is there; Question 46 on the next page explores this.

Question 35

If $\cos^{-1}\alpha + \cos^{-1}\beta + \cos^{-1}\gamma = 3\pi$, then $\alpha(\beta + \gamma) + \beta(\gamma + \alpha) + \gamma(\alpha + \beta)$ equals

Solution. Each $\cos^{-1}$ is at most $\pi$, so three of them can total $3\pi$ only if each equals $\pi$. That forces $\alpha = \beta = \gamma = \cos\pi = -1$. The expression is $2(\alpha\beta + \beta\gamma + \gamma\alpha) = 2(1 + 1 + 1) = 6$.

Question 36

The number of real solutions of the equation $\sqrt{1 + \cos2x} = \sqrt2\cos^{-1}(\cos x)$ in $\left[\dfrac\pi2, \pi\right]$ is

Solution. The left side is $\sqrt{2\cos^2x} = \sqrt2|\cos x|$, and on $\left[\frac\pi2, \pi\right]$, $\cos x \le 0$, so it is $-\sqrt2\cos x$. The right side is $\sqrt2x$, because $x$ is already in $[0, \pi]$. The equation becomes $-\cos x = x$.

On this interval $-\cos x$ runs from $0$ to $1$, while $x$ is at least $\frac\pi2 \approx 1.57$. The two sides never meet.

Question 37

If $\cos^{-1}x > \sin^{-1}x$, then

Solution. Replace $\cos^{-1}x$ by $\frac\pi2 – \sin^{-1}x$:

$$\frac\pi2 – \sin^{-1}x > \sin^{-1}x \iff \sin^{-1}x < \frac\pi4 \iff x < \sin\frac\pi4 = \frac{1}{\sqrt2}$$

($\sin$ is increasing on the branch, so the inequality keeps its direction). Together with the domain $[-1, 1]$ this is $-1 \le x < \frac{1}{\sqrt2}$.

The printed key says (A), and that is the range where the inequality goes the other way. Test $x = 0$: $\cos^{-1}0 = \frac\pi2$ and $\sin^{-1}0 = 0$, so $\cos^{-1}x > \sin^{-1}x$ holds, yet $0$ is not in (A). Test $x = 1$, which (A) includes: $\cos^{-1}1 = 0 < \frac\pi2 = \sin^{-1}1$. Option (B) is true for every $x$ it lists but leaves out $[-1, 0)$, where the inequality also holds.

Common mistakes

• Mixing up the branches. $\sin^{-1}$, $\tan^{-1}$ and $\mathrm{cosec}^{-1}$ live around $0$; $\cos^{-1}$, $\cot^{-1}$ and $\sec^{-1}$ live in $[0, \pi]$. Questions 20 and 21 offer the wrong family as distractors.
• Cancelling before reducing. $\cos^{-1}\left(\cos\frac{3\pi}{2}\right)$ is not $\frac{3\pi}{2}$ (Question 28). Find the value inside first or reduce the angle into the branch.
• Forgetting that a range decides which angle. In Questions 26 and 36 the cosine is fixed, but the answer depends on knowing which interval the angle must lie in.
• Reversing an inequality by accident. In Question 37 the step from $\sin^{-1}x < \frac\pi4$ to $x < \frac{1}{\sqrt2}$ keeps the direction because $\sin$ is increasing on $\left[-\frac\pi2, \frac\pi2\right]$. The printed key’s answer is exactly the reversed range.
• Doubling inside the function. $2\tan^{-1}0.75$ is not $\tan^{-1}1.5$ (Question 27). Use the double-angle formula on the angle, not the input.

Practise next

• Exemplar Short and Long Answer Q1–19, the proofs and evaluations from this unit, including the derivations of the addition formulas.
• Exemplar Fill in the Blanks and True or False Q38–55, the rest of the unit.
• NCERT textbook Exercise 2.1 for principal values, the skill behind Questions 20 to 23 and 28.
• NCERT textbook Exercise 2.2 for evaluating expressions like those in Questions 29, 32 and 33.