NCERT Class 12 Mathematics — Inverse Trigonometric Functions, Miscellaneous Exercise on Chapter 2. All 14 questions solved.

Fourteen questions in three groups: two evaluations where the inner angle sits outside the principal branch, eight identities to prove, and two equations to solve, followed by two multiple-choice questions.

The identities in questions 3 to 7 all follow one recipe. Convert every term to the same inverse function — usually $\tan^{-1}$ — using a right-angled triangle, then apply the addition formula:

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

Key insight. A statement like $\sin^{-1}\tfrac{8}{17}$ is just an angle in a right-angled triangle with opposite side $8$ and hypotenuse $17$. The third side is $15$, so the same angle is $\tan^{-1}\tfrac{8}{15}$ and $\cos^{-1}\tfrac{15}{17}$. Every question from 3 to 7 becomes routine once you convert every term into whichever inverse function makes the identity you want to use apply.

Look for Pythagorean triples: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$ and $(16,63,65)$ all appear in this exercise.

Find the value of the following.

Question 1

$\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)$

Solution. The branch of $\cos^{-1}$ is $[0, \pi]$, and $\tfrac{13\pi}{6}$ is well outside it — it is more than a full turn. Reduce using the $2\pi$ period:

$$\cos\frac{13\pi}{6} = \cos\left(2\pi + \frac{\pi}{6}\right) = \cos\frac{\pi}{6}$$

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

Question 2

$\tan^{-1}\left(\tan\dfrac{7\pi}{6}\right)$

Solution. Tangent has period $\pi$, not $2\pi$, so subtract $\pi$ once:

$$\tan\frac{7\pi}{6} = \tan\left(\frac{7\pi}{6} – \pi\right) = \tan\frac{\pi}{6}$$

and $\tfrac{\pi}{6}$ lies in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$.

Prove that:

Question 3

$2\sin^{-1}\dfrac35 = \tan^{-1}\dfrac{24}{7}$

Solution. Let $\theta = \sin^{-1}\tfrac35$. In a right-angled triangle with opposite $3$ and hypotenuse $5$, the adjacent side is $4$, so $\tan\theta = \tfrac34$ and $\theta = \tan^{-1}\tfrac34$.

Now use the double-angle form of $\tan^{-1}$:

$$2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2} \qquad (|x| < 1)$$

With $x = \tfrac34$:

$$\frac{2 \cdot \frac34}{1 – \frac{9}{16}} = \frac{\frac32}{\frac{7}{16}} = \frac{24}{7}$$

Since $\tfrac34 < 1$, the formula applies with no $\pi$ correction, and $2\sin^{-1}\tfrac35 = \tan^{-1}\tfrac{24}{7}$.

Question 4

$\sin^{-1}\dfrac{8}{17} + \sin^{-1}\dfrac35 = \tan^{-1}\dfrac{77}{36}$

Solution. Convert both terms to arctangents. The triple $(8, 15, 17)$ gives $\sin^{-1}\tfrac{8}{17} = \tan^{-1}\tfrac{8}{15}$; the triple $(3, 4, 5)$ gives $\sin^{-1}\tfrac35 = \tan^{-1}\tfrac34$.

The product $\tfrac{8}{15} \cdot \tfrac34 = \tfrac{2}{5} < 1$, so the addition formula applies directly:

$$\tan^{-1}\frac{\frac{8}{15} + \frac34}{1 – \frac25} = \tan^{-1}\frac{\frac{77}{60}}{\frac35} = \tan^{-1}\frac{77}{36}$$

Question 5

$\cos^{-1}\dfrac45 + \cos^{-1}\dfrac{12}{13} = \cos^{-1}\dfrac{33}{65}$

Solution. Here the natural identity is the one for cosines:

$$\cos^{-1}x + \cos^{-1}y = \cos^{-1}\left(xy – \sqrt{1-x^2}\sqrt{1-y^2}\right)$$

With $x = \tfrac45$ we get $\sqrt{1-x^2} = \tfrac35$, and with $y = \tfrac{12}{13}$ we get $\sqrt{1-y^2} = \tfrac{5}{13}$:

$$\frac45 \cdot \frac{12}{13} – \frac35 \cdot \frac{5}{13} = \frac{48}{65} – \frac{15}{65} = \frac{33}{65}$$

Both angles are acute and their sum is less than $\pi$, so the result stays in $\cos^{-1}$’s branch.

Question 6

$\cos^{-1}\dfrac{12}{13} + \sin^{-1}\dfrac35 = \sin^{-1}\dfrac{56}{65}$

Solution. Convert the first term to a sine so both sides speak the same language: the triple $(5, 12, 13)$ gives $\cos^{-1}\tfrac{12}{13} = \sin^{-1}\tfrac{5}{13}$.

Then apply

$$\sin^{-1}x + \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$$

with $x = \tfrac{5}{13}$ and $y = \tfrac35$:

$$\frac{5}{13} \cdot \frac45 + \frac35 \cdot \frac{12}{13} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$$

Question 7

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

Solution. Convert the right-hand side to arctangents. The triple $(5, 12, 13)$ gives $\sin^{-1}\tfrac{5}{13} = \tan^{-1}\tfrac{5}{12}$, and $(3, 4, 5)$ gives $\cos^{-1}\tfrac35 = \tan^{-1}\tfrac43$.

Now $\tfrac{5}{12} \cdot \tfrac43 = \tfrac59 < 1$, so:

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

The triple $(16, 63, 65)$ confirms the result is a genuine triangle angle.

Prove that:

Question 8

$\tan^{-1}\sqrt{x} = \dfrac12\cos^{-1}\left(\dfrac{1-x}{1+x}\right)$, $x \in [0, 1]$

Solution. Put $\sqrt{x} = \tan\theta$, so $x = \tan^2\theta$ and $\theta = \tan^{-1}\sqrt{x}$. Then

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

using the standard double-angle form. Hence $\cos^{-1}\left(\tfrac{1-x}{1+x}\right) = 2\theta$, provided $2\theta \in [0, \pi]$.

The condition $x \in [0, 1]$ gives $\theta \in \left[0, \tfrac{\pi}{4}\right]$, so $2\theta \in \left[0, \tfrac{\pi}{2}\right]$ — comfortably inside the branch. Dividing by $2$ gives the result.

Question 9

$\cot^{-1}\left(\dfrac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} – \sqrt{1-\sin x}}\right) = \dfrac{x}{2}$, $x \in \left(0, \tfrac{\pi}{4}\right)$

Solution. The two roots simplify once you write $1$ as $\sin^2\tfrac{x}{2} + \cos^2\tfrac{x}{2}$ and $\sin x$ as $2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$, making each a perfect square:

$$1 \pm \sin x = \left(\cos\frac{x}{2} \pm \sin\frac{x}{2}\right)^2$$

For $x \in \left(0, \tfrac{\pi}{4}\right)$ we have $\tfrac{x}{2} < \tfrac{\pi}{8}$, so $\cos\tfrac{x}{2} > \sin\tfrac{x}{2} > 0$ and both square roots are taken positive without a modulus:

$$\sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}, \qquad \sqrt{1-\sin x} = \cos\frac{x}{2} – \sin\frac{x}{2}$$

The numerator becomes $2\cos\tfrac{x}{2}$ and the denominator $2\sin\tfrac{x}{2}$, so the fraction is $\cot\tfrac{x}{2}$ and the expression is $\tfrac{x}{2}$, which lies in $(0, \pi)$ as $\cot^{-1}$ requires.

Question 10

$\tan^{-1}\left(\dfrac{\sqrt{1+x} – \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \dfrac{\pi}{4} – \dfrac12\cos^{-1}x$, $-\dfrac{1}{\sqrt2} \le x \le 1$  [Hint: Put $x = \cos 2\theta$]

Solution. With $x = \cos 2\theta$, the two roots become half-angle expressions:

$$1 + x = 2\cos^2\theta, \qquad 1 – x = 2\sin^2\theta$$

so $\sqrt{1+x} = \sqrt2\cos\theta$ and $\sqrt{1-x} = \sqrt2\sin\theta$, both positive for $\theta \in \left[0, \tfrac{3\pi}{8}\right]$ — which is what the stated range of $x$ delivers. The fraction becomes

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

dividing through by $\cos\theta$ and recognising $1 = \tan\tfrac{\pi}{4}$, exactly as in Exercise 2.2 question 5. Since $\theta = \tfrac12\cos^{-1}x$, the whole expression is $\tfrac{\pi}{4} – \tfrac12\cos^{-1}x$.

Solve the following equations.

Question 11

$2\tan^{-1}(\cos x) = \tan^{-1}(2\operatorname{cosec} x)$

Solution. Apply the double-angle identity to the left side:

$$2\tan^{-1}(\cos x) = \tan^{-1}\left(\frac{2\cos x}{1 – \cos^2 x}\right) = \tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right)$$

Since $\tan^{-1}$ is one-to-one, the arguments must be equal:

$$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}$$

Multiplying through by $\sin^2 x$ (permissible because $\operatorname{cosec} x$ already requires $\sin x \ne 0$) gives $2\cos x = 2\sin x$, so $\tan x = 1$.

Question 12

$\tan^{-1}\dfrac{1-x}{1+x} = \dfrac12\tan^{-1}x$, $(x > 0)$

Solution. The left side is a disguised difference. Since $\tfrac{1-x}{1+x} = \tfrac{1 – x}{1 + 1 \cdot x}$, the subtraction formula gives

$$\tan^{-1}\frac{1-x}{1+x} = \tan^{-1}1 – \tan^{-1}x = \frac{\pi}{4} – \tan^{-1}x$$

Substituting:

$$\frac{\pi}{4} – \tan^{-1}x = \frac12\tan^{-1}x \quad\Longrightarrow\quad \frac{\pi}{4} = \frac32\tan^{-1}x$$

So $\tan^{-1}x = \tfrac{\pi}{6}$.

Question 13

$\sin\left(\tan^{-1}x\right)$, $|x| < 1$ is equal to

Solution. Let $\theta = \tan^{-1}x$, so $\tan\theta = \tfrac{x}{1}$. Draw the right-angled triangle: opposite $x$, adjacent $1$, hypotenuse $\sqrt{1+x^2}$. Then

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$$

Options (A) and (B) contain $\sqrt{1-x^2}$, which belongs to $\sin^{-1}$ and $\cos^{-1}$ problems, not $\tan^{-1}$ ones; option (C) is $\cos\left(\tan^{-1}x\right)$.

Question 14

$\sin^{-1}(1-x) – 2\sin^{-1}x = \dfrac{\pi}{2}$, then $x$ is equal to

Solution. Rearrange so one inverse sine stands alone, then take the sine of both sides:

$$\sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1}x$$

$$1 – x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}x\right) = \cos\left(2\sin^{-1}x\right) = 1 – 2x^2$$

using $\cos(2\theta) = 1 – 2\sin^2\theta$ with $\theta = \sin^{-1}x$. So $2x^2 = x$, giving $x = 0$ or $x = \tfrac12$.

Both roots must now be tested in the original equation, because squaring-type steps can introduce extras. At $x = 0$: $\sin^{-1}1 – 0 = \tfrac{\pi}{2}$ ✓. At $x = \tfrac12$: $\sin^{-1}\tfrac12 – 2\sin^{-1}\tfrac12 = -\tfrac{\pi}{6}$, which is not $\tfrac{\pi}{2}$ ✗.

So $x = \tfrac12$ is extraneous and option (A) is the trap for anyone who stops at the quadratic.

Common mistakes

• Not checking the roots of an inverse trigonometric equation. Question 14 is built entirely on this: the algebra produces two values and only one satisfies the equation.
• Reducing an angle with the wrong period. Cosine repeats every $2\pi$ (question 1), tangent every $\pi$ (question 2). Using $2\pi$ on the tangent question leaves you outside the branch.
• Applying the addition formula when $xy > 1$. Every use of it in questions 4 and 7 was preceded by checking the product. Where $xy > 1$ the formula needs an added $\pi$.
• Dropping the modulus when simplifying a square root. In questions 9 and 10 the roots are taken positive only because the stated domain makes the inside positive. Without that check, $\sqrt{(\cos\tfrac{x}{2} – \sin\tfrac{x}{2})^2}$ is the modulus, not the bracket.
• Converting between inverse functions carelessly. $\sin^{-1}\tfrac{8}{17} = \tan^{-1}\tfrac{8}{15}$, not $\tan^{-1}\tfrac{8}{17}$. The third side of the triangle has to be computed.

Practise next

• Exercise 2.2 — the identities and substitutions these proofs rely on.
• Exercise 2.1 — the principal value branches that decide questions 1, 2 and 11.