NCERT Exemplar Solutions Class 12 Maths Inverse Trigonometric Functions — Short and Long Answer Q1–19

NCERT Exemplar Problems, Class 12 Mathematics — Inverse Trigonometric Functions, Short Answer questions 1 to 11 and Long Answer questions 12 to 19. All 19 questions solved.

Nineteen questions in two kinds. The first is “undo the function” — $\tan^{-1}(\tan\theta)$, $\cos^{-1}(\cos\theta)$ — where the whole skill is pulling the angle back into the principal value branch before cancelling (Questions 1, 2, 4, 5, 13). The second is the addition and double-angle formulas for $\tan^{-1}$, which drive most of the proofs (3, 6, 9, 10, 14 to 17, 19).

A note on the syllabus. The rationalised NCERT textbook still has a section on properties of inverse trigonometric functions, but it no longer lists the formulas $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x+y}{1-xy}$ and $2\tan^{-1}x = \tan^{-1}\dfrac{2x}{1-x^2}$ that this exemplar uses throughout. JEE Main and Advanced still test them, and each one is a single line from $\tan(A + B)$, so the solutions below derive them where they first appear rather than quote them.

Two questions are misprinted. Question 9 states its answer without the $n$ it refers to, and the interval in Question 13 is slightly too wide for the printed answer. Both are explained where they occur.

Key insight. $\sin^{-1}(\sin\theta) = \theta$ only when $\theta$ already lies in $\left[-\frac\pi2, \frac\pi2\right]$, and the same goes for every inverse function and its own branch. Before cancelling, move the angle into the branch using the periodicity and symmetry of the outer function. Nearly every wrong answer in this unit is a cancellation done one step too early.

Question 1

Find the value of $\tan^{-1}\left(\tan\dfrac{5\pi}{6}\right) + \cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)$.

Solution. $\frac{5\pi}{6}$ is outside $\left(-\frac\pi2, \frac\pi2\right)$, the branch of $\tan^{-1}$. Since $\tan$ has period $\pi$, $\tan\frac{5\pi}{6} = \tan\left(\frac{5\pi}{6} – \pi\right) = \tan\left(-\frac\pi6\right)$, and $-\frac\pi6$ is in the branch. So the first term is $-\frac\pi6$.

$\frac{13\pi}{6}$ is outside $[0, \pi]$, the branch of $\cos^{-1}$. Since $\cos$ has period $2\pi$, $\cos\frac{13\pi}{6} = \cos\frac\pi6$, so the second term is $\frac\pi6$.

Cancelling straight away would give $\frac{5\pi}{6} + \frac{13\pi}{6} = 3\pi$, which is not even in the possible range of the sum.

Question 2

Evaluate $\cos\left[\cos^{-1}\left(\dfrac{-\sqrt3}{2}\right) + \dfrac\pi6\right]$.

Solution. The angle in $[0, \pi]$ whose cosine is $-\frac{\sqrt3}{2}$ is $\frac{5\pi}{6}$, so $\cos^{-1}\left(-\frac{\sqrt3}{2}\right) = \frac{5\pi}{6}$. The bracket is $\frac{5\pi}{6} + \frac\pi6 = \pi$, and $\cos\pi = -1$.

Question 3

Prove that $\cot\left(\dfrac\pi4 – 2\cot^{-1}3\right) = 7$.

Solution. Since $3 > 0$, $\cot^{-1}3 = \tan^{-1}\frac13$. Let $\theta = \tan^{-1}\frac13$, so $\tan\theta = \frac13$. The double-angle formula for tangent gives

$$\tan 2\theta = \frac{2\tan\theta}{1 – \tan^2\theta} = \frac{\frac23}{1 – \frac19} = \frac{\frac23}{\frac89} = \frac34.$$

Now $\tan\left(\frac\pi4 – 2\theta\right) = \dfrac{1 – \tan2\theta}{1 + \tan2\theta} = \dfrac{1 – \frac34}{1 + \frac34} = \dfrac{\frac14}{\frac74} = \dfrac17$, so $\cot\left(\frac\pi4 – 2\theta\right) = 7$.

Question 4

Find the value of $\tan^{-1}\left(-\dfrac{1}{\sqrt3}\right) + \cot^{-1}\left(\dfrac{1}{\sqrt3}\right) + \tan^{-1}\left(\sin\left(\dfrac{-\pi}{2}\right)\right)$.

Solution. Take the three terms one at a time, each in its own branch.

• $\tan^{-1}\left(-\frac{1}{\sqrt3}\right) = -\frac\pi6$, since $\tan^{-1}$ is odd and $\tan\frac\pi6 = \frac{1}{\sqrt3}$.
• $\cot^{-1}\frac{1}{\sqrt3} = \frac\pi3$, since $\cot\frac\pi3 = \frac{1}{\sqrt3}$ and $\frac\pi3 \in (0, \pi)$.
• $\sin\left(-\frac\pi2\right) = -1$, and $\tan^{-1}(-1) = -\frac\pi4$.

Adding: $-\frac\pi6 + \frac\pi3 – \frac\pi4 = \frac{-2 + 4 – 3}{12}\pi$.

Question 5

Find the value of $\tan^{-1}\left(\tan\dfrac{2\pi}{3}\right)$.

Solution. $\frac{2\pi}{3}$ lies outside $\left(-\frac\pi2, \frac\pi2\right)$. Subtracting one period, $\tan\frac{2\pi}{3} = \tan\left(\frac{2\pi}{3} – \pi\right) = \tan\left(-\frac\pi3\right)$, and $-\frac\pi3$ is inside the branch.

Question 6

Show that $2\tan^{-1}(-3) = \dfrac{-\pi}{2} + \tan^{-1}\left(\dfrac{-4}{3}\right)$.

Solution. $\tan^{-1}$ is odd, so the left side is $-2\tan^{-1}3$. The double-angle formula $2\tan^{-1}x = \tan^{-1}\frac{2x}{1 – x^2}$ holds only for $|x| < 1$, and here $x = 3$, so it needs care. Let $\theta = \tan^{-1}3$. Then $\theta \in \left(\frac\pi4, \frac\pi2\right)$, so $2\theta \in \left(\frac\pi2, \pi\right)$, and

$$\tan2\theta = \frac{2 \cdot 3}{1 – 9} = -\frac34.$$

An angle in $\left(\frac\pi2, \pi\right)$ with tangent $-\frac34$ is $\pi – \tan^{-1}\frac34$. So $2\tan^{-1}3 = \pi – \tan^{-1}\frac34$ and the left side is $-\pi + \tan^{-1}\frac34$.

For the right side, use $\tan^{-1}\frac43 = \frac\pi2 – \tan^{-1}\frac34$ (the two acute angles of a 3–4–5 triangle are complementary):

$$-\frac\pi2 + \tan^{-1}\left(-\frac43\right) = -\frac\pi2 – \left(\frac\pi2 – \tan^{-1}\frac34\right) = -\pi + \tan^{-1}\frac34.$$

Question 7

Find the real solutions of the equation $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \dfrac\pi2$.

Solution. Let $\alpha = \tan^{-1}\sqrt{x(x+1)}$, which lies in $\left[0, \frac\pi2\right)$. The equation says $\sin^{-1}\sqrt{x^2 + x + 1} = \frac\pi2 – \alpha$, so

$$\sqrt{x^2 + x + 1} = \sin\left(\frac\pi2 – \alpha\right) = \cos\alpha = \frac{1}{\sqrt{1 + \tan^2\alpha}} = \frac{1}{\sqrt{1 + x^2 + x}}.$$

Multiplying across, $x^2 + x + 1 = 1$, so $x(x + 1) = 0$ and $x = 0$ or $x = -1$.

Check both, because the square roots need $x(x+1) \ge 0$ and $x^2 + x + 1 \le 1$: at either value $x(x+1) = 0$ and $x^2 + x + 1 = 1$, giving $\tan^{-1}0 + \sin^{-1}1 = 0 + \frac\pi2$.

Question 8

Find the value of the expression $\sin\left(2\tan^{-1}\dfrac13\right) + \cos\left(\tan^{-1}2\sqrt2\right)$.

Solution. For the first term use $\sin2\theta = \dfrac{2\tan\theta}{1 + \tan^2\theta}$ with $\tan\theta = \frac13$:

$$\sin\left(2\tan^{-1}\frac13\right) = \frac{\frac23}{1 + \frac19} = \frac{\frac23}{\frac{10}{9}} = \frac35.$$

For the second, if $\tan\phi = 2\sqrt2$ with $\phi$ acute, then $\sec^2\phi = 1 + 8 = 9$, so $\cos\phi = \frac13$.

$\frac35 + \frac13 = \frac{9 + 5}{15}$.

Question 9

If $2\tan^{-1}(\cos\theta) = \tan^{-1}(2\,\mathrm{cosec}\,\theta)$, then show that $\theta = n\pi + \dfrac\pi4$, where $n$ is any integer.

The book prints “show that $\theta = \frac\pi4$, where $n$ is any integer” — the $n\pi$ has dropped out of the statement. The general solution is $n\pi + \frac\pi4$, proved below.

Solution. Since $|\cos\theta| \le 1$, the double-angle formula applies on the left (for $|\cos\theta| < 1$; $\cos\theta = \pm1$ would make $\mathrm{cosec}\,\theta$ undefined anyway):

$$2\tan^{-1}(\cos\theta) = \tan^{-1}\frac{2\cos\theta}{1 – \cos^2\theta} = \tan^{-1}\frac{2\cos\theta}{\sin^2\theta}.$$

$\tan^{-1}$ is one-one, so the equation becomes

$$\frac{2\cos\theta}{\sin^2\theta} = \frac{2}{\sin\theta} \implies \cos\theta = \sin\theta \implies \tan\theta = 1 \implies \theta = n\pi + \frac\pi4.$$

Both families work. For $\theta = \frac\pi4$, $2\tan^{-1}\frac{1}{\sqrt2}$ and $\tan^{-1}2\sqrt2$ are equal (tangent of the first is $\frac{\sqrt2}{1 – \frac12} = 2\sqrt2$). For $\theta = \frac{5\pi}{4}$ both sides change sign together.

This is Question 11 of the textbook’s Miscellaneous Exercise, written with $\theta$ instead of $x$.

Question 10

Show that $\cos\left(2\tan^{-1}\dfrac17\right) = \sin\left(4\tan^{-1}\dfrac13\right)$.

Solution. Left side, with $\cos2\theta = \dfrac{1 – \tan^2\theta}{1 + \tan^2\theta}$ and $\tan\theta = \frac17$:

$$\cos\left(2\tan^{-1}\frac17\right) = \frac{1 – \frac{1}{49}}{1 + \frac{1}{49}} = \frac{48}{50} = \frac{24}{25}.$$

Right side: Question 3 showed $2\tan^{-1}\frac13 = \tan^{-1}\frac34$, so $4\tan^{-1}\frac13 = 2\tan^{-1}\frac34$, and with $\sin2\phi = \dfrac{2\tan\phi}{1 + \tan^2\phi}$:

$$\sin\left(2\tan^{-1}\frac34\right) = \frac{\frac32}{1 + \frac{9}{16}} = \frac{\frac32}{\frac{25}{16}} = \frac{24}{25}.$$

Question 11

Solve the equation $\cos\left(\tan^{-1}x\right) = \sin\left(\cot^{-1}\dfrac34\right)$.

Solution. If $\tan\theta = x$ with $\theta \in \left(-\frac\pi2, \frac\pi2\right)$, then $\cos\theta > 0$ and $\cos\theta = \dfrac{1}{\sqrt{1 + x^2}}$. If $\cot\phi = \frac34$ with $\phi \in (0, \pi)$, then $\phi$ is acute, the triangle is 3–4–5, and $\sin\phi = \frac45$. So

$$\frac{1}{\sqrt{1 + x^2}} = \frac45 \implies 1 + x^2 = \frac{25}{16} \implies x^2 = \frac{9}{16}.$$

Both signs are valid: $\cos(\tan^{-1}x)$ is even in $x$.

Question 12

Prove that $\tan^{-1}\left(\dfrac{\sqrt{1 + x^2} + \sqrt{1 – x^2}}{\sqrt{1 + x^2} – \sqrt{1 – x^2}}\right) = \dfrac\pi4 + \dfrac12\cos^{-1}x^2$.

Solution. The two roots $\sqrt{1 \pm x^2}$ suggest putting $x^2 = \cos2\theta$, because $1 + \cos2\theta = 2\cos^2\theta$ and $1 – \cos2\theta = 2\sin^2\theta$. For $0 < x^2 \le 1$ take $2\theta = \cos^{-1}x^2 \in \left[0, \frac\pi2\right)$, so $\theta \in \left[0, \frac\pi4\right)$ and $\sin\theta, \cos\theta \ge 0$. Then $\sqrt{1 + x^2} = \sqrt2\cos\theta$ and $\sqrt{1 – x^2} = \sqrt2\sin\theta$, and

$$\frac{\sqrt2\cos\theta + \sqrt2\sin\theta}{\sqrt2\cos\theta – \sqrt2\sin\theta} = \frac{1 + \tan\theta}{1 – \tan\theta} = \tan\left(\frac\pi4 + \theta\right).$$

Since $\frac\pi4 + \theta \in \left[\frac\pi4, \frac\pi2\right)$ is in the branch of $\tan^{-1}$, the left side is $\frac\pi4 + \theta = \frac\pi4 + \frac12\cos^{-1}x^2$.

Question 13

Find the simplified form of $\cos^{-1}\left(\dfrac35\cos x + \dfrac45\sin x\right)$, where $x \in \left[\dfrac{-3\pi}{4}, \dfrac\pi4\right]$.

Solution. $\frac35$ and $\frac45$ are the cosine and sine of one angle: let $\alpha = \tan^{-1}\frac43$, so $\cos\alpha = \frac35$ and $\sin\alpha = \frac45$. Then

$$\frac35\cos x + \frac45\sin x = \cos x\cos\alpha + \sin x\sin\alpha = \cos(x – \alpha).$$

$\cos^{-1}(\cos y) = -y$ when $y \in [-\pi, 0]$. With $\alpha \approx 0.927$, $y = x – \alpha$ is in $[-\pi, 0]$ exactly when $\alpha – \pi \le x \le \alpha$, which covers $\left[\alpha – \pi, \frac\pi4\right]$. On that range

The interval in the question is slightly too wide for this answer. $\alpha – \pi \approx -2.214$, but $-\frac{3\pi}{4} \approx -2.356$. On the sliver $-\frac{3\pi}{4} \le x < \tan^{-1}\frac43 – \pi$, the angle $x – \alpha$ is below $-\pi$, and the value is $2\pi + x – \tan^{-1}\frac43$ instead. At $x = -\frac{3\pi}{4}$, for instance, the expression is about $3.000$, while $\tan^{-1}\frac43 – x \approx 3.283$, which is more than $\pi$ and cannot be a $\cos^{-1}$ value. The printed key gives only $\tan^{-1}\frac43 – x$; that is the intended answer and is right on almost all of the interval.

Question 14

Prove that $\sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac35 = \sin^{-1}\dfrac{77}{85}$.

Solution. Let $A = \sin^{-1}\frac{8}{17}$ and $B = \sin^{-1}\frac35$, both acute. Then $\cos A = \frac{15}{17}$ (an 8–15–17 triangle) and $\cos B = \frac45$. So

$$\sin(A + B) = \frac{8}{17}\cdot\frac45 + \frac{15}{17}\cdot\frac35 = \frac{32 + 45}{85} = \frac{77}{85}.$$

That alone does not finish it: $\sin^{-1}$ returns an angle in $\left[-\frac\pi2, \frac\pi2\right]$, so we need $A + B$ to be acute. $\cos(A + B) = \frac{15}{17}\cdot\frac45 – \frac{8}{17}\cdot\frac35 = \frac{60 – 24}{85} = \frac{36}{85} > 0$, so it is, and $A + B = \sin^{-1}\frac{77}{85}$.

Question 15

Show that $\sin^{-1}\dfrac{5}{13} + \cos^{-1}\dfrac35 = \tan^{-1}\dfrac{63}{16}$.

Solution. Turn both into $\tan^{-1}$, using the right-angled triangles. $\sin^{-1}\frac{5}{13}$ is acute with opposite 5, hypotenuse 13, adjacent 12, so it equals $\tan^{-1}\frac{5}{12}$. $\cos^{-1}\frac35$ is acute with adjacent 3, hypotenuse 5, opposite 4, so it equals $\tan^{-1}\frac43$.

The addition formula comes from $\tan(A + B) = \dfrac{\tan A + \tan B}{1 – \tan A\tan B}$; it gives $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x + y}{1 – xy}$ whenever $xy < 1$. Here $xy = \frac{5}{12}\cdot\frac43 = \frac59 < 1$:

$$\tan^{-1}\frac{5}{12} + \tan^{-1}\frac43 = \tan^{-1}\frac{\frac{5}{12} + \frac{16}{12}}{1 – \frac59} = \tan^{-1}\frac{\frac{21}{12}}{\frac49} = \tan^{-1}\frac{63}{16}.$$

Question 16

Prove that $\tan^{-1}\dfrac14 + \tan^{-1}\dfrac29 = \sin^{-1}\dfrac{1}{\sqrt5}$.

Solution. The product $\frac14\cdot\frac29 = \frac{1}{18} < 1$, so the addition formula applies:

$$\tan^{-1}\frac14 + \tan^{-1}\frac29 = \tan^{-1}\frac{\frac14 + \frac29}{1 – \frac{1}{18}} = \tan^{-1}\frac{\frac{17}{36}}{\frac{34}{36}} = \tan^{-1}\frac12.$$

An acute angle with tangent $\frac12$ sits in a triangle with sides 1, 2 and hypotenuse $\sqrt5$, so its sine is $\frac{1}{\sqrt5}$: $\tan^{-1}\frac12 = \sin^{-1}\frac{1}{\sqrt5}$.

Question 17

Find the value of $4\tan^{-1}\dfrac15 – \tan^{-1}\dfrac{1}{239}$.

Solution. Double twice. $2\tan^{-1}\frac15 = \tan^{-1}\dfrac{\frac25}{1 – \frac{1}{25}} = \tan^{-1}\frac{5}{12}$, and $4\tan^{-1}\frac15 = 2\tan^{-1}\frac{5}{12} = \tan^{-1}\dfrac{\frac{10}{12}}{1 – \frac{25}{144}} = \tan^{-1}\dfrac{120}{119}$ (both doublings are valid because $\frac15$ and $\frac{5}{12}$ are below $1$).

Then subtract, using $\tan^{-1}x – \tan^{-1}y = \tan^{-1}\dfrac{x – y}{1 + xy}$ for $xy > -1$:

$$\tan^{-1}\frac{120}{119} – \tan^{-1}\frac{1}{239} = \tan^{-1}\frac{\frac{120 \cdot 239 – 119}{119 \cdot 239}}{\frac{119 \cdot 239 + 120}{119 \cdot 239}} = \tan^{-1}\frac{28561}{28561} = \tan^{-1}1.$$

This is Machin’s formula, which in 1706 gave the first hundred digits of $\pi$.

Question 18

Show that $\tan\left(\dfrac12\sin^{-1}\dfrac34\right) = \dfrac{4 – \sqrt7}{3}$ and justify why the other value $\dfrac{4 + \sqrt7}{3}$ is ignored.

Solution. Let $\theta = \sin^{-1}\frac34$, so $\theta \in \left(0, \frac\pi2\right)$ and $\cos\theta = \sqrt{1 – \frac{9}{16}} = \frac{\sqrt7}{4}$. With $t = \tan\frac\theta2$, $\sin\theta = \dfrac{2t}{1 + t^2}$, so

$$\frac{2t}{1 + t^2} = \frac34 \implies 3t^2 – 8t + 3 = 0 \implies t = \frac{8 \pm \sqrt{64 – 36}}{6} = \frac{4 \pm \sqrt7}{3}.$$

Which root? $\theta \in \left(0, \frac\pi2\right)$ means $\frac\theta2 \in \left(0, \frac\pi4\right)$, so $0 < \tan\frac\theta2 < 1$. Now $\frac{4 + \sqrt7}{3} \approx 2.22 > 1$ is ruled out, while $\frac{4 – \sqrt7}{3} \approx 0.45$ fits. (Directly: $\tan\frac\theta2 = \dfrac{1 – \cos\theta}{\sin\theta} = \dfrac{1 – \frac{\sqrt7}{4}}{\frac34} = \dfrac{4 – \sqrt7}{3}$.)

The quadratic cannot tell $\frac\theta2$ from the angle $\frac\pi2 – \frac\theta2$, which has the same $\sin\theta$ for its double; the product of the roots is $1$ for exactly that reason.

Question 19

If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$, then evaluate

$$\tan\left[\tan^{-1}\left(\frac{d}{1 + a_1a_2}\right) + \tan^{-1}\left(\frac{d}{1 + a_2a_3}\right) + \tan^{-1}\left(\frac{d}{1 + a_3a_4}\right) + \cdots + \tan^{-1}\left(\frac{d}{1 + a_{n-1}a_n}\right)\right].$$

Solution. Write the numerator $d$ as $a_{k+1} – a_k$. Then each term has the shape of the subtraction formula:

$$\tan^{-1}\frac{a_{k+1} – a_k}{1 + a_ka_{k+1}} = \tan^{-1}a_{k+1} – \tan^{-1}a_k$$

(valid when $1 + a_ka_{k+1} > 0$, the usual assumption here). The sum telescopes — each $\tan^{-1}a_k$ appears once with each sign — leaving $\tan^{-1}a_n – \tan^{-1}a_1$. Taking the tangent with $\tan(A – B)$:

Common mistakes

• Cancelling $\tan^{-1}(\tan\theta)$ or $\cos^{-1}(\cos\theta)$ without checking the branch. Questions 1, 5 and 13 are built to catch this. Shift the angle by a period, or reflect it, until it sits in the branch, then cancel.
• Using the double-angle formula outside $|x| < 1$. $2\tan^{-1}x = \tan^{-1}\frac{2x}{1 – x^2}$ fails for $x = 3$ (Question 6): the left side is beyond $\frac\pi2$ and the right side cannot be.
• Using $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 – xy}$ without checking $xy < 1$. In Questions 15 to 17 the products are all below $1$, which is why the formula applies; it is worth writing that check down every time.
• Stopping at $\sin(A + B) = \frac{77}{85}$ in Question 14. Two angles share each sine value; the sign of $\cos(A + B)$ decides whether $A + B$ is the principal one.
• Keeping both roots of a quadratic in Question 18. The range of the half-angle, not the algebra, decides which one belongs.

Practise next

• Exemplar Objective Type Q20–37, the eighteen multiple-choice questions from this unit, one of which the key gets wrong.
• Exemplar Fill in the Blanks and True or False Q38–55, short principal-value checks.
• NCERT textbook Exercise 2.2 for simplifying with a substitution, as in Question 12.
• NCERT textbook Miscellaneous Exercise on Inverse Trigonometric Functions, which contains Question 9 almost word for word.