NCERT Exemplar Solutions Class 12 Maths Inverse Trigonometric Functions — Fill in the Blanks and True or False Q38–55

NCERT Exemplar Problems, Class 12 Mathematics — Inverse Trigonometric Functions, Fill in the Blanks questions 38 to 48 and True or False questions 49 to 55. All 18 questions solved.

Eighteen short questions, most answerable in one line once you know the branches. The blanks are principal values (38, 39, 42, 43), the two complementary identities (40, 44, 45), the domain of $\sec^{-1}$ (41), and the conditions under which the standard formulas hold (46 to 48). The true-or-false statements test what an inverse trigonometric function actually is: why the domain has to be restricted, and what “principal value” means.

Key insight. An inverse trigonometric function returns exactly one angle, from a fixed interval. Every blank on this page is “find the angle in that interval”, and every true-or-false statement is about why that interval is needed.

Question 38

The principal value of $\cos^{-1}\left(-\dfrac12\right)$ is $\rule{2cm}{0.4pt}$.

Solution. $\cos\frac\pi3 = \frac12$, and on $[0, \pi]$ the angle with cosine $-\frac12$ is its supplement, $\pi – \frac\pi3$. In general $\cos^{-1}(-x) = \pi – \cos^{-1}x$.

Question 39

The value of $\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right)$ is $\rule{2cm}{0.4pt}$.

Solution. $\frac{3\pi}{5}$ is past $\frac\pi2$, outside the branch. Since $\sin(\pi – \theta) = \sin\theta$, $\sin\frac{3\pi}{5} = \sin\frac{2\pi}{5}$, and $\frac{2\pi}{5}$ is inside $\left[-\frac\pi2, \frac\pi2\right]$.

Question 40

If $\cos\left(\tan^{-1}x + \cot^{-1}\sqrt3\right) = 0$, then the value of $x$ is $\rule{2cm}{0.4pt}$.

Solution. $\cot^{-1}\sqrt3 = \frac\pi6$. The bracket $\tan^{-1}x + \frac\pi6$ lies in $\left(-\frac\pi3, \frac{2\pi}{3}\right)$, and the only angle there with cosine $0$ is $\frac\pi2$. So $\tan^{-1}x = \frac\pi3$ and $x = \sqrt3$. (Equivalently: $\tan^{-1}x + \cot^{-1}x = \frac\pi2$, so the bracket is $\frac\pi2$ exactly when $x = \sqrt3$.)

Question 41

The set of values of $\sec^{-1}\left(\dfrac12\right)$ is $\rule{2cm}{0.4pt}$.

Solution. $\sec\theta = \frac{1}{\cos\theta}$ is never between $-1$ and $1$, because $|\cos\theta| \le 1$. So $\frac12$ is outside the domain $\mathbb R – (-1, 1)$ of $\sec^{-1}$, and there is no value.

Question 42

The principal value of $\tan^{-1}\sqrt3$ is $\rule{2cm}{0.4pt}$.

Solution. $\tan\frac\pi3 = \sqrt3$ and $\frac\pi3 \in \left(-\frac\pi2, \frac\pi2\right)$.

Question 43

The value of $\cos^{-1}\left(\cos\dfrac{14\pi}{3}\right)$ is $\rule{2cm}{0.4pt}$.

Solution. Remove two whole periods: $\frac{14\pi}{3} – 4\pi = \frac{2\pi}{3}$, so $\cos\frac{14\pi}{3} = \cos\frac{2\pi}{3}$, and $\frac{2\pi}{3} \in [0, \pi]$.

Question 44

The value of $\cos\left(\sin^{-1}x + \cos^{-1}x\right)$, $|x| \le 1$, is $\rule{2cm}{0.4pt}$.

Solution. $\sin^{-1}x + \cos^{-1}x = \frac\pi2$ for every $x \in [-1, 1]$, and $\cos\frac\pi2 = 0$.

Question 45

The value of the expression $\tan\left(\dfrac{\sin^{-1}x + \cos^{-1}x}{2}\right)$, when $x = \dfrac{\sqrt3}{2}$, is $\rule{2cm}{0.4pt}$.

Solution. The numerator is $\frac\pi2$ whatever $x$ is, so the expression is $\tan\frac\pi4 = 1$. The value $x = \frac{\sqrt3}{2}$ is a distraction; working out $\frac\pi3 + \frac\pi6$ gives the same thing.

Question 46

If $y = 2\tan^{-1}x + \sin^{-1}\left(\dfrac{2x}{1 + x^2}\right)$ for all $x$, then $\rule{1cm}{0.4pt} < y < \rule{1cm}{0.4pt}$.

Solution. Split by the size of $x$, with $x = \tan\theta$, $\theta \in \left(-\frac\pi2, \frac\pi2\right)$, so that $\dfrac{2x}{1 + x^2} = \sin2\theta$.

• $|x| \le 1$: $2\theta \in \left[-\frac\pi2, \frac\pi2\right]$, so $\sin^{-1}(\sin2\theta) = 2\theta$ and $y = 4\tan^{-1}x$, which runs over $[-\pi, \pi]$.
• $x > 1$: $2\theta \in \left(\frac\pi2, \pi\right)$, so $\sin^{-1}(\sin2\theta) = \pi – 2\theta$ and $y = 2\theta + \pi – 2\theta = \pi$.
• $x < -1$: by the same reasoning $y = -\pi$.

So $y$ takes exactly the values $-\pi \le y \le \pi$, and reaches $\pm\pi$ (in fact it equals $\pi$ for every $x \ge 1$).

The blank, as printed, cannot be filled tightly. Because $y = \pi$ is attained, there is no number $b$ with $y < b$ that is also the least upper bound; any $b > \pi$ works and none is best. The printed key gives $-2\pi$ and $2\pi$, which is true but loose. The useful statement, and the one worth remembering, is $-\pi \le y \le \pi$, with $y$ constant at $\pm\pi$ outside $[-1, 1]$.

Question 47

The result $\tan^{-1}x – \tan^{-1}y = \tan^{-1}\left(\dfrac{x – y}{1 + xy}\right)$ is true when the value of $xy$ is $\rule{2cm}{0.4pt}$.

Solution. The formula comes from $\tan(A – B) = \dfrac{\tan A – \tan B}{1 + \tan A\tan B}$, which is correct for any angles; the issue is whether $A – B$ lands in $\left(-\frac\pi2, \frac\pi2\right)$, the range of $\tan^{-1}$. When $xy > -1$, the denominator $1 + xy$ is positive, so $\cos(A – B) = \cos A\cos B\,(1 + xy)$ is positive and $A – B$ is in the branch. When $xy < -1$, $A – B$ falls outside it and the formula is off by $\pm\pi$ (for example $x = 2$, $y = -1$).

Question 48

The value of $\cot^{-1}(-x)$ for all $x \in \mathbb R$ in terms of $\cot^{-1}x$ is $\rule{2cm}{0.4pt}$.

Solution. The branch of $\cot^{-1}$ is $(0, \pi)$, which is symmetric about $\frac\pi2$, not about $0$. If $\cot\theta = x$ with $\theta \in (0, \pi)$, then $\cot(\pi – \theta) = -x$ and $\pi – \theta$ is also in $(0, \pi)$. So $\cot^{-1}(-x) = \pi – \cot^{-1}x$, the same pattern as $\cos^{-1}$.

Question 49

State True or False: All trigonometric functions have inverse over their respective domains.

Solution. A function has an inverse only if it is one-one, and no trigonometric function is one-one on its whole domain: $\sin0 = \sin\pi = 0$, for a start. That is why we restrict each to a branch first.

Question 50

State True or False: The value of the expression $(\cos^{-1}x)^2$ is equal to $\sec^2x$.

Solution. $\cos^{-1}x$ is the inverse function (an angle), not the reciprocal $\frac{1}{\cos x} = \sec x$. At $x = 1$, $(\cos^{-1}1)^2 = 0$, while $\sec^21 \approx 3.43$.

Question 51

State True or False: The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Solution. Any interval on which the function is one-one and covers its full range gives a valid inverse. For $\sin$, $\left[\frac\pi2, \frac{3\pi}{2}\right]$ works as well as $\left[-\frac\pi2, \frac\pi2\right]$. The principal branch is the agreed choice, not the only possible one.

Question 52

State True or False: The least numerical value, either positive or negative, of angle $\theta$ is called principal value of the inverse trigonometric function.

Solution. This is the older definition the exemplar uses: among all angles with the given trigonometric value, the principal value is the one of least absolute value. For $\sin^{-1}$, $\tan^{-1}$ and $\cos^{-1}$ it picks out the branch value, provided a tie is broken in favour of the positive angle — for $\cos^{-1}\left(-\frac12\right)$, both $\frac{2\pi}{3}$ and $-\frac{2\pi}{3}$ have least absolute value, and the principal value is $\frac{2\pi}{3}$. The key marks the statement True, and that is the intended answer.

Treat it as a rule of thumb, not a definition. It breaks for $\cot^{-1}$ of a negative number: the angles with cotangent $-1$ include $-\frac\pi4$ and $\frac{3\pi}{4}$, and the numerically least is $-\frac\pi4$, but with the branch $(0, \pi)$ the principal value is $\cot^{-1}(-1) = \frac{3\pi}{4}$. The current textbook defines the principal value as the value in the principal value branch, which never has this problem.

Question 53

State True or False: The graph of inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging $x$ and $y$ axes.

Solution. The graph of any inverse function is the reflection of the original in the line $y = x$, which is what swapping the axes does. For an inverse trigonometric function you reflect the branch, not the whole periodic curve.

Question 54

State True or False: The minimum value of $n$ for which $\tan^{-1}\dfrac{n}{\pi} > \dfrac\pi4$, $n \in \mathbb N$, is valid is $5$.

Solution. $\tan^{-1}$ is increasing, so $\tan^{-1}\frac n\pi > \frac\pi4 = \tan^{-1}1$ exactly when $\frac n\pi > 1$, i.e. $n > \pi \approx 3.14$. The least natural number is $4$, not $5$.

Question 55

State True or False: The principal value of $\sin^{-1}\left[\cos\left(\sin^{-1}\dfrac12\right)\right]$ is $\dfrac\pi3$.

Solution. Work from the inside. $\sin^{-1}\frac12 = \frac\pi6$, $\cos\frac\pi6 = \frac{\sqrt3}{2}$, and $\sin^{-1}\frac{\sqrt3}{2} = \frac\pi3$.

Common mistakes

• Reading $\cos^{-1}x$ as $\frac{1}{\cos x}$. The $-1$ marks the inverse function, not a power (Question 50). The reciprocal is $\sec x$, or $(\cos x)^{-1}$ with brackets.
• Using $\cot^{-1}(-x) = -\cot^{-1}x$. That is the rule for $\sin^{-1}$ and $\tan^{-1}$, whose branches are symmetric about $0$. $\cot^{-1}$ and $\cos^{-1}$ follow $\pi$ minus (Questions 38, 48).
• Giving a value for $\sec^{-1}\frac12$. The input is outside the domain, so the answer is the empty set (Question 41).
• Forgetting the condition on a formula. The subtraction formula needs $xy > -1$ (Question 47), and $2\tan^{-1}x + \sin^{-1}\frac{2x}{1 + x^2} = 4\tan^{-1}x$ needs $|x| \le 1$ (Question 46).
• Rounding a strict inequality. $n > \pi$ gives $n = 4$, not $5$ and not $3$ (Question 54).

Practise next

• Exemplar Short and Long Answer Q1–19, where the addition and double-angle formulas used here are derived and applied.
• Exemplar Objective Type Q20–37, eighteen multiple-choice questions, one of which the key gets wrong.
• NCERT textbook Exercise 2.1 for principal values like those in Questions 38, 39, 42 and 43.
• NCERT textbook Miscellaneous Exercise on Inverse Trigonometric Functions for more expressions of the $\cos^{-1}\left(\cos\theta\right)$ type.