33. cos−1(cos7π6)is equal to(A)7π6 (B)5π6 (C)π3 (D)π6

Cos-1 cos 7π6 (M) As 7π6 [0, π] ;principal value branch of cos-1 cos-1 (cos7π6) = cos-1 (cos2x−7π6) {? cos (2π−θ)=cosθ} = cos-1 (sis12π−7π6) =cos−1cos5π65π6∈ [0, π] = 5π6 So, option B is correct

15. tan−1(xy)−tan−1x−yx=y is equal to (A) π2 (B) π3 (C) π4 (D) 3π4

We know that. sin 3θ =3 sin θ 4sin3θ (identity). (E) Let x = sinθ. Then, sin −1x=θ . We have, Sin3 (sin −1x) = 3x−4x3 3sin −1x =sin-1 (3x−4x3) Hence proved.

42. tan−16316=sin−1513+cos−135

Let sin−1513=x and cos−135=y. Then, sinx=513and cos=35 So, tanx=sinxcosxandtany=sinycosy =5/312/1 =4/53/5. =512 =43. Using tan(x+y)=lanx+lany1−tanxtany. tan (sin−1513+cos−135)=512+431−512×43= 5×3+4×1212×312×3−5×412×3 =15+4836−20=6316 ⇒sin−1513+cos−135=tan−16316. Hence proved.

tan−1(tan3π4). (M). As 3π4∉(−π2,π2); principal value branch of tan -1we can write, tan−1(tan3π4)=tan−1tan4π−π4 =tan−1(tan4π4−π4) =tan−1(tanπ−π4) =tan−1(−tanπ4){?tanis(−)weinI−1 quadent } =−tan−1(tan14) {?tan−1(−x)=−tan−1x} = −π4

Class 12 Maths Chapter 2 Inverse Trigonometric Functions NCERT Solutions

Updated on Jul 31, 2025 12:16 IST

By Pallavi Pathak, Assistant Manager Content

Inverse Trigonometric Functions in Class 12 is important for students preparing for the CBSE Board exams and other entrance tests, such as JEE Mains. The concepts covered in the Inverse Trigonometric Functions Class 12 chapter are also used in science and engineering. The inverse trigonometric functions are also important in calculus.
Inverse Trigonometry Class 12 covers the basic concepts and the properties of the inverse trigonometric functions. Students must thoroughly study the NCERT solutions to understand the concepts, improve their problem-solving skills, and be confident to appear in their examinations. The solution will help students in solving even the most complex problems of the chapter.
Explore here the comprehensive NCERT solutions of the Class 11 and Class 12 for all three subjects - Maths, Physics, and Chemistry. The solutions are given in a step-by-step format.

Table of contents

Key Concepts of Trigonometric Functions – Class 12 at a Glance
Class 12 Maths Ch 2 Inverse Trigonometric Functions – NCERT Solutions PDF
Inverse Trigonometric Function Exercise-wise NCERT Solutions
Class 12 Math Chapter 2 Inverse Trigonometric Function Exercise 2.1 Solutions
NCERT Class 12 Math Chapter 2: Key Topics, Weightage
Important Formulas of Inverse Trigonometry Class 12
NCERT Solutions for Inverse Trigonometric Functions- FAQs

Key Concepts of Trigonometric Functions – Class 12 at a Glance

Find below an overview of the Inverse Trigonometric Functions Class 12:
Here is a table showing the domains and ranges of inverse trigonometric functions:

Functions	Domain	Range
$y = \sin^{- 1} x$	[–1, 1]	$[\frac{- π}{2}, \frac{π}{2}]$
$y = \cos^{- 1} (x)$	[–1, 1]	[0, π]
$y = \csc^{- 1} (x)$	R – (–1,1)	$[\frac{- π}{2}, \frac{π}{2}] - {0}$
$y = \sec^{- 1} (x)$	R – (–1, 1)	$[0, π] - {\frac{π}{2}}$
$y = \tan^{- 1} (x)$	R	$(\frac{- π}{2}, \frac{π}{2})$
$y = \cot^{- 1} (x)$	R	(0, π)

Get the Class 12 Maths NCERT Solutions here. It will provide all the important topics for concept clarity, chapter-wise weightage information and free PDFs.

Class 12 Maths Ch 2 Inverse Trigonometric Functions – NCERT Solutions PDF

Shiksha provides the free Inverse Trigonometry Class 12 PDF link below. Students must download it and practice from it to score well in the CBSE Board exams and other competitive exams.

Class 12 Math Chapter 2 Inverse Trigonometric Function Solution: Free PDF Download

Inverse Trigonometric Function Exercise-wise NCERT Solutions

Inverse Trigonometric Function is an essential topic for building foundational trigonometric concepts. Chapter 2 Class 12 Math Inverse Trigonometric Function deals with the properties, domains, and ranges of inverse trigonometric functions. The ITF concepts also play a vital role in calculus, Matrices, Determinants, and other important physics and Math concepts and real-world applications. Students can check the exercise-wise Chapter 2 ITF math solutions below;

Class 12 Math Chapter 2 Inverse Trigonometric Function Exercise 2.1 Solutions

Chapter 2 Inverse Trigonometric Function Exercise 2.1 focuses on problems finding principal values, range, domain, and co-domain. Chapter 2 ITF Exercise 2.1 includes 14 Questions (12 Short Answers, 2 MCQs) primarily based on finding the values of inverse trigonometric functions. Students can check the complete the exercise 2.1 solutions below;

Inverse Trigonometric Function Exercise 2.1 Solutions

Find the principal values of the following:

Q1.

s i n^{- 1} (- \frac{1}{2})

A.1. Let sin⁻¹ $(\frac{- 1}{2})$ =y. Then, sin y=- $\frac{- 1}{2}$

We know that the range of principal value branch of sin⁻¹ is $- \frac{π}{2},$ $- \frac{π}{2}$

and sin⁻¹ $(\frac{- 1}{2})$ =−sin⁻¹ $\frac{- 1}{2}$ (sin (-x) = -sin x)

= $(- \frac{π}{6})$ = sin y (as sin $\frac{π}{6}$ = $\frac{1}{2}$ )

Principal value of sin⁻¹ $(- \frac{1}{2})$ is $(- \frac{π}{6})$

Commonly asked questions

33. $\begin{array}{l} c o s^{- 1} (c o s \frac{7 π}{6}) i s e q u a l t o \\ (A) \frac{7 π}{6} (B) \frac{5 π}{6} (C) \frac{π}{3} (D) \frac{π}{6} \end{array}$

Cos^-1 cos $\frac{7 π}{6}$

(M) As $\frac{7 π}{6}$ [0, $π$ ] ;principal value branch of cos^-1

cos^-1 $(c o s \frac{7 π}{6})$ = cos^-1 $(c o s 2 x - \frac{7 π}{6})$ ${? c o s (2 π - θ) = c o s θ}$

= cos^-1 $(s i s \frac{12 π - 7 π}{6})$

$= c o s^{- 1} c o s \frac{5 π}{6} \frac{5 π}{6} \in [0, π]$

= $\frac{5 π}{6}$

So, option B is correct

15. $t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} \frac{x - y}{x = y} i s e q u a l t o$

$(A) \frac{π}{2} (B) \frac{π}{3} (C) \frac{π}{4} (D) \frac{3 π}{4}$

We know that.

sin 3θ =3 sin θ 4sin³θ (identity).

(E) Let x = sinθ. Then, sin ⁻¹x=θ . We have,

Sin3 (sin ⁻¹x) = 3x−4x³

3sin ⁻¹x =sin^-1 (3x−4x³)

Hence proved.

42. $t a n^{- 1} \frac{63}{16} = s i n^{- 1} \frac{5}{13} + c o s^{- 1} \frac{3}{5}$

Let $s i n^{- 1} \frac{5}{13} = x and c o s^{- 1} \frac{3}{5} = y .$

Then, $s i n x = \frac{5}{13} a n d c o s = \frac{3}{5}$

So, $t a n x = \frac{s i n x}{c o s x} a n d t a n y = \frac{s i n y}{c o s y}$

$= \frac{5 / 3}{12 / 1} = \frac{4 / 5}{3 / 5} .$

$= \frac{5}{12} = \frac{4}{3} .$

Using $t a n (x + y) = \frac{l a n x + l a n y}{1 - t a n x t a n y .}$

tan $(s i n^{- 1} \frac{5}{13} + c o s^{- 1} \frac{3}{5}) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}} =$ $\frac{\frac{5 \times 3 + 4 \times 12}{12 \times 3}}{\frac{12 \times 3 - 5 \times 4}{12 \times 3}}$

$= \frac{15 + 48}{36 - 20} = \frac{63}{16}$

$\Rightarrow s i n^{- 1} \frac{5}{13} + c o s^{- 1} \frac{3}{5} = t a n^{- 1} \frac{63}{16} .$

Hence proved.

44. Kindly Consider the following

Kindly go through the solution

31. $t a n^{- 1} (t a n \frac{3 π}{4})$

$t a n^{- 1} (t a n \frac{3 π}{4}) .$

(M). As $\frac{3 π}{4} \notin (- \frac{π}{2}, \frac{π}{2});$ principal value branch of tan ^-1we can write,

$t a n^{- 1} (t a n \frac{3 π}{4}) = t a n^{- 1} t a n \frac{4 π - π}{4}$ $= t a n^{- 1} (t a n \frac{4 π}{4} - \frac{π}{4})$

$= t a n^{- 1} (t a n π - \frac{π}{4})$

$= t a n^{- 1} (- t a n \frac{π}{4}) {? t a n is(-)wein I^{- 1}$ quadent }

$= - t a n^{- 1} (t a n \frac{1}{4}) {? t a n^{- 1} (- x) = - t a n^{- 1} x}$

= $- \frac{π}{4}$

38. Kindly Consider the following

Kindly go through the solution

Solve the following equations:

48. $2 t a n^{- 1} (c o s x) = t a n^{- 1} (2 c o s e c x)$

Given,

$(M) 2 t a n^{- 1} (c o s x) = t a n^{- 1} (2 c o s e c x)$

$\Rightarrow t a n^{- 1} \frac{2 c o s x}{1 - c o s^{2} x} = t a n^{- 1} \frac{2}{s i n x}$ ${\begin{cases} u s i n g . \\ t a n^{- 1} 2 x = \frac{2 x}{1 - x^{2}} \end{cases}}$

$\Rightarrow \frac{2 c o s x}{s i n^{2} x} = \frac{2}{s i n x}$ ${? 1 - c o s^{2} x = s i n^{2} x \Rightarrow 1 = s i n^{2} x + c o s^{2} x}$

$\Rightarrow \frac{c o s x}{s i n x} = 1$

$\Rightarrow c o t x = c o t \frac{x}{4}$

$\Rightarrow x = \frac{π}{4} .$

11. $t a n^{- 1} (1) + c o s^{- 1} - \frac{1}{2} + s i n^{- 1} - \frac{1}{2}$

tan⁻¹ (1) + cos⁻¹ $(\frac{1}{2})$ + sin ⁻¹ $(\frac{- 1}{2})$

39. $2 s i n^{- 1} \frac{8}{17} + s i n^{- 1} \frac{3}{5} = t a n^{- 1} \frac{77}{36}$

Let $s i n^{- 1} \frac{8}{17} = x s i n^{- 1} \frac{3}{5} = y .$

(N. then, $s i n x = \frac{8}{17} s i n y = \frac{3}{5} .$

So, $t a n x = \frac{s i n x}{c o s x}$ and $t a n y = \frac{s i n y}{c o s y}$

$= \frac{8 / 17}{15 / 17} = \frac{3 / 4}{4 / 5}$

$= \frac{8}{15} = \frac{3}{4}$

Using. $tan (x + y) = \frac{t a n x + t a n y}{1 - t a n x t a n y}$

$t a n (s i n^{- 1} \frac{8}{17} + s i n^{- 1} \frac{3}{5}) = \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}}$

tan $(s i n^{- 1} \frac{8}{17} + s i n^{- 1} \frac{3}{5} = \frac{\frac{8 \times 4 + 3 \times 15}{15 \times 4}}{\frac{15 \times 4 - 8 \times 3}{15 \times 4}} = \frac{32 + 45}{60 - 24} \Rightarrow s i n$

sin^-1 $\frac{8}{17} + s i n^{- 1} \frac{3}{5} =$ tan^-1 $\frac{77}{36}$

Hence proved.

46. Kindly Consider the following

Kindly go through the solution

17. $t a n^{- 1} \frac{2}{11} + t a n^{- 1} \frac{7}{24} = t a n^{- 1} \frac{1}{2}$

L.H.S = tan ^-1 $\frac{2}{11}$ + tan ^-1 $\frac{7}{24}$

Using tan ^-1x+ tan ^-1y= tan ^-1 $\frac{x + y}{1 - x y}$ , xy<1

L.H.S =tan ^-1 $\frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \times \frac{7}{24}}$ = tan ^-1 $\frac{\frac{2 \times 24 + 7 \times 2}{11 \times 24}}{\frac{11 \times 24 - 7 \times 2}{11 \times 24}}$

tan ^-1 $\frac{48 + 14}{264 - 14}$ = tan ^-1 $\frac{125}{250}$ = tan ^-1 $\frac{1}{2}$ = R.H.S

20. Kindly Consider the following

Kindly go through the solution

52. $\begin{array}{l} t a n^{- 1} (\frac{x}{y}) - t a n^{- 1} \frac{x - y}{x = y} i s e q u a l t o \\ (A) \frac{π}{2} (B) \frac{π}{3} (C) \frac{π}{4} (D) \frac{3 π}{4} \end{array}$

$t a n^{- 1} \frac{x}{y} - t a n^{- 1} \frac{x - y}{x + y} {- t a n^{- 1} x - t a n^{- 1} y = \frac{x - y}{1 + x y}}$

$= t a n^{- 1} {\frac{(\frac{x}{y}) - (\frac{x - y}{x + y})}{1 + \frac{x}{y} \cdot (\frac{x - y}{x + y})}}$

$= t a n^{- 1} {\frac{\frac{x (x + y) - (x - y) \cdot y}{y (x + y)}}{\frac{y (x + y) + x (x - y)}{y (x + y)}}}$

$= t a n^{- 1} (\frac{x^{2} + x y - x y + y^{2}}{x y + y^{2} - x^{2} - x y})$

$= t a n^{- 1} \frac{x^{2} + y^{2}}{x^{2} + y^{2}}$

$= t a n^{- 1} 1$

$= t a n^{- 1} (t a n \frac{π}{4})$

$= \frac{π}{4} .$

Option C is correct.

28. $I f s i n (s i n^{1} \frac{1}{5} + c o s^{1} x) = 1, t h e n f i n d t h e v a l u e o f x$

$s i n (s i n^{- 1} \frac{1}{5} + c o s^{- 1} x) = 1 .$

(E) $\Rightarrow s i n (s i n^{- 1} \frac{1}{5} + c o s^{- 1} x) = s i n \frac{π}{2} {? s i n \frac{π}{2} = 1}$

$\Rightarrow s i n^{- 1} \frac{1}{5} + c o s^{- 1} x = s i n^{- 1} (s i n \frac{π}{2}) = \frac{π}{2}$

$\Rightarrow c o s^{- 1} x = \frac{π}{2} - s i n^{- 1} \frac{1}{5}$

$\Rightarrow c o s^{- 1} x = c o s^{- 1} \frac{1}{5}$ ${? \frac{π}{2} = s i n^{- 1} x + c o s^{- 1} x}$

$= \frac{π}{2} - s i n^{- 1} \frac{1}{5} = c o s^{- 1} \frac{1}{5}$ ${\begin{cases} \Rightarrow \frac{π}{2} - s i n^{- 1} x = c o s^{- 1} x \\ x = \frac{1}{5} \\ \Rightarrow \frac{π}{2} - s i n^{- 1} \frac{1}{5} = c o s^{- 1} \frac{1}{5} \end{cases} .$

= x = $\frac{1}{5} .$

Find the values of each of the following:

25. $t a n^{- 1} [2 c o s (2 s i n^{1} \frac{1}{2})]$

$t a n^{- 1} [2 c o s (2 s i n^{- 1} \frac{1}{2})] = t a n^{- 1} [2 c o s (2 s i n^{- 1} s i n \frac{π}{6})]$

(E) $= t a n^{- 1} [2 c o s (2 \times \frac{π}{6})]$

$= t a n^{- 1} [2 c o s \frac{π}{3}]$

$= t a n^{- 1} 2 \times \frac{1}{2}$

$= t a n^{- 1} 1$

$= t a n^{- 1} (t a n \frac{π}{4})$

$= \frac{π}{4}$

36. $c o s^{- 1} (c o s \frac{13 π}{6})$

Cos^-1 $(c o s \frac{13 π}{6 .})$ = $c o s^{- 1}$ $(c o s \frac{12 π + π}{6})$

= $c o s^{- 1}$ $(c o s \frac{12 π}{6} + \frac{π}{6})$

= $c o s^{- 1}$ $[c o s^{- 1} (2 π + \frac{π}{6})]$ {Øcor2 $π$ +=ØcosØ}

$= c o s^{- 1} (c o s \frac{π}{6})$

$= \frac{π}{6}$ ∈ [0, x]

24. Kindly Consider the following

Kindly go through the solution

45. Kindly Consider the following

Kindly go through the solution

27. $t a n \frac{1}{2} [s i n^{- 1} \frac{2 x}{1 + x^{2}} + c o s^{- 1} \frac{1 - y^{2}}{1 + y^{2}}], | x | < 1, y > 0 a n d x y < 1$

$t a n \frac{1}{2} [s i n^{- 1} \frac{2 x}{1 + x^{2}} + c o s^{- 1} \frac{1 - y^{2}}{1 + y^{2}}], | x | < 1, y >$

(M) Let x = tanØ . then tan^-1x= 0 and y = tan ω then tan^-1y = ω. we have,

tan $\frac{1}{2}$ ${sin^{-1} \frac{2 t a n θ}{1 + t a n^{2} θ} + \frac{1 - t a n^{2} ω}{1 + t a n^{2} ω}}$

$= t a n \frac{1}{2} {s i n^{- 1} (s i n 2 θ) + c o s^{- 1} (c o s 2 ω)}$ ${∴ s i n 2 θ \frac{2 t a n θ}{1 + t a n^{2} θ} c o s 2 θ = \frac{1 - t a n^{2} θ}{1 + t a n^{2} θ} t a n x + y = \frac{t a n x + t a n y}{1 - t a n x t a n y}}$

$= t a n \frac{1}{2} {2 θ + 2 ω} = t a n (θ + ω)$

$= \frac{t a n θ + t a n ω}{1 - t a n θ t a n ω}$

= $\frac{x = y}{1 - x y}$

49. $t a n^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} t a n^{- 1} x, (x > 0)$

Given, (M)

$t a n^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} t a n^{- 1} x$

$x = t a n θ.Then θ = t a n^{- 1} x Bo we have,$

$t a n^{- 1} (\frac{1 - t a n θ}{1 + t a n θ}) = \frac{1}{2} t a n^{- 1} (t a n θ)$

$\Rightarrow t a n^{- 1} (\frac{t a n \frac{π}{4} - t a n θ}{1 + t a n \frac{x}{4} t a n θ}) = \frac{1}{2} θ {? t a n \frac{π}{4} = 1} .$

$\Rightarrow t a n^{- 1} {t a n (\frac{x}{4} - θ)} = \frac{θ}{2} {\frac{? t a n x - t a n y}{1 + t a n x t a n} = t a n (x - y)$

$\Rightarrow \frac{π}{4} - θ = \frac{θ}{2}$

$\Rightarrow \frac{θ}{2} + θ = \frac{x}{4}$

$\Rightarrow \frac{3 θ}{2} = \frac{π}{4}$

$\Rightarrow θ = \frac{π}{4} \times \frac{2}{3} = \frac{π}{6}$

23. Kindly Consider the following

Kindly go through the solution

50. Kindly Consider the following

Kindly go through the solution

1. $s i n^{- 1} (- \frac{1}{2})$

Let sin⁻¹ $(\frac{- 1}{2})$ =y. Then, sin y=- $\frac{- 1}{2}$

We know that the range of principal value branch of sin⁻¹ is $- \frac{π}{2},$ $- \frac{π}{2}$

and sin⁻¹ $(\frac{- 1}{2})$ =−sin⁻¹ $\frac{- 1}{2}$ (sin (-x) = -sin x)

= $(- \frac{π}{6})$ = sin y (as sin
$\frac{π}{6}$
= $\frac{1}{2}$ )

Principal value of sin⁻¹ $(- \frac{1}{2})$ is $(- \frac{π}{6})$

12. $c o s^{- 1} \frac{1}{2} + 2 s i n^{- 1} \frac{1}{2}$

cos⁻¹ $\frac{1}{2}$ + 2Sin⁻¹ $\frac{1}{2}$ = cos⁻¹ $(c o s \frac{π}{3})$ + 2×Sin⁻¹ $(s i n \frac{π}{6})$

= $\frac{π}{3} + 2 \times \frac{π}{6}$

= $\frac{π}{3} + \frac{π}{3}$

= $\frac{2 π}{3}$

13. $I f s i n^{- 1} x = y, t h e n$

$\begin{array}{l} (A) 0 \leq y \leq π (B) - \frac{π}{2} \leq y \leq \frac{π}{2} \\ (C) 0 < y < π (D) - \frac{π}{2} < y < \frac{π}{2} \end{array}$

Given, Sin⁻¹x=y.

(E) We know that the principal value branch of Sin⁻¹ is

$[\frac{- π}{2}, \frac{π}{2}]$ Hence, $\frac{- π}{2}$ ≤ y ≤ $\frac{π}{2}$

Option B is correct.

5. $c o s^{- 1} (- \frac{1}{2})$

Let cos ^-1 $(\frac{- 1}{2})$ =y Then cos y = $\frac{- 1}{2}$ = − cos
$\frac{π}{3}$ $\frac{}{}$ cos $(π - \frac{x}{3})$

= cos $\frac{3 π - π}{3}$

= cos $\frac{2 π}{3}$

(E) We know that the range of principal value

branch of cos⁻¹ is [0, $π$ ] and cos $\frac{2 x}{3}$ = $\frac{- 1}{2}$

Principal value of cos−1 $(\frac{- 1}{2})$ is $\frac{2 x}{3}$

26. $c o t (t a n^{- 1} a + c o t^{- 1} a)$

cot $(t a n^{- 1} a + c o t^{- 1} a) = c o t \frac{π}{2} {? t a n^{- 1} x + c o t^{- 1} x = \frac{π}{2}}$

(E) $= 0$

29. $I f t a n^{- 1} \frac{x - 1}{x - 2} + t a n^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}, t h e n f i n d t h e v a l u e o f x$

$t a n^{- 1} \frac{x - 1}{x - 2} + t a n^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}$

(E)Using $t a n^{- 1} x + t a n^{- 1} y = t a n^{- 1} \frac{x + y}{1 - x y} .$

$\Rightarrow t a n^{- 1} \frac{\frac{(x - 1)}{(x - 2)} + (\frac{x + 1}{x + 2})}{1 - \frac{(x - 1)}{(x - 2)} \times (\frac{x + 1}{x + 2})} = \frac{π}{4} .$

$\Rightarrow \frac{\frac{(x - 1) (x + 2) - 1 (x + 1) (x - 2)}{(x - 2) (x + 2)}}{\frac{(x - 2) (x + 2) - (x - 1) (x + 1)}{(x - 2) (x + 2)}} = t a n \frac{π}{4}$

$\Rightarrow \frac{(x - 1) (x + 2) + (x + 1) (x - 2)}{(x - 2) (x + 2) - (x - 1) (x + 1)} = 1 .$

$\Rightarrow \frac{x^{2} + 2 x - x - 2 + x^{2} - 2 x + x - 2}{(x^{2} - 4) - (x^{2} - 1^{2})} = 1 .$ ${\begin{matrix} ? (a - b) (a + b) = & a^{2} - b^{2} \end{matrix}}$

$\Rightarrow \frac{2 x^{2} - 4}{x^{2} - 4 - x^{2} + 1} = 1 \Rightarrow \frac{2 x^{2} - 4}{- 3} = 1 \Rightarrow 2 x^{2} - 4 = - 3$

Find the values of each of the expressions in Exercises 16 to 18.

30. $s i n^{- 1} (s i n \frac{2 π}{3})$

$s i n^{- 1} (s i n \frac{2 π}{3}) .$

(M) As $\frac{2 π}{3} \notin [- \frac{π}{2}, \frac{π}{2}]$ principal value branch of sin-1 we can write,

$s i n^{- 1} (s i n \frac{3 π - π}{3}) = s i n^{- 1} (s i n \frac{3 π}{3} - \frac{π}{3}) = s i n^{- 1} (s i n π - \frac{π}{3})$

$= s i n^{- 1} (s i n \frac{π}{3}) {? s i nis (+)ve in ind quadrat$
$^{}}$

$= \frac{π}{3} .$

16. $3 c o s^{- 1} x = c o s^{- 1} (4 x^{3} - 3 x), x \in [\frac{1}{2}, 1]$

We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3 (cos^-1x) = 4x³-3x

3cos^-1x = cos^-1 (4x³- 3x)

Hence Proved

18. $2 t a n^{- 1} \frac{1}{2} + t a n^{- 1} \frac{1}{7} = t a n^{- 1} \frac{31}{17}$

L.H.S= 2 tan ^-1 $\frac{1}{2}$ + tan ^-1 $\frac{1}{7}$

(E) Using2 tan ^-1x= tan ^-1 $\frac{2 a}{1 - x^{2}}$ we can write.

L.H.S = tan ^-1 $\frac{2 \times \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{1 \frac{- 1}{4}}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{4 \frac{- 1}{4}}$ + tan ^-1 $\frac{1}{7}$ = tan ^-1 $\frac{4}{3}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \times \frac{1}{7}}$ { Ø tan -1 x + tan -1 y = tan - 1 $\frac{x + y}{1 - x y}$ }

= tan ^-1 $\frac{\frac{7 \times 4 + 1 \times 3}{3 \times 7}}{\frac{3 \times 7 - 4 \times 1}{3 \times 7}}$

= tan -1 $\frac{28 + 3}{21 - 4}$ = tan ^-1 $\frac{31}{17}$ = R H S

34. $\begin{array}{l} s i n^{- 1} (\frac{π}{3} - s i n^{- 1} (- \frac{1}{2})) i s e q u a l t o \\ (A) \frac{1}{2} (B) \frac{1}{3} (C) \frac{1}{4} (D) 1 \end{array}$

= sin $[\frac{π}{3} + s i n^{- 1} (s i n \frac{π}{6})]$

= sin $[\frac{π}{3} + \frac{π}{6}]$ = sin $[\frac{2 π + π}{6}]$ = sin $(\frac{3 π}{6})$

= sin $\frac{π}{2} = 1$

14. $t a n^{- 1}$ √3 $- s e c^{- 1} (- 2) i s e q u a l t o$

$(A) π (B) - \frac{π}{3} (C) \frac{π}{3} (D) \frac{2 π}{3}$

=tan⁻¹ $(\frac{t a n π}{3})$ − $π$ + sec⁻¹ $(s e c \frac{π}{3})$

= $\frac{π}{3} - π + \frac{π}{3}$

= $\frac{π - 3 π + π}{3}$

= $\frac{- π}{3} .$

Option B is correct.

22. $t a n^{- 1} (\frac{c o s x - s i n x}{c o s x + s i n x}), 0 \frac{- π}{4} < x < \frac{3 π}{4}$

Given tan ^-1 $\frac{c o s x - s i n x}{c o s x + s i n x}$ , $\frac{- π}{4},$ x< $\frac{3 x}{4}$

(M) Dividing numerator & denominator cos x we get,

tan ^-1 $\frac{\frac{c o s x - s i n x}{c o s x}}{\frac{c o s x + s i n x}{c o s x}} = t a n^{- 1} \frac{\frac{c o s x}{c o s x} - \frac{s i n x}{c o s x}}{\frac{c o s x}{c o s x} + \frac{s i n x}{c o s x}} = t a n^{- 1} \frac{1 - t a n x}{1 + t a n x .}$

We know that $t a n \frac{π}{4} = 1$ = 1 so,

$= t a n^{- 1} \frac{t a n \frac{π}{4} - t a n x}{1 + t a n \frac{π}{4} t a n x} = t a n^{- 1} [t a n (\frac{π}{4} - x)]$ ${? t a n (x - y) = \frac{t a n x - t a n y}{1 + t a n x \cdot t a n y}}$

$= \frac{π}{4} - x$ .

32. $t a n (s i n^{- 1} \frac{3}{5} + c o {t^{-}}^{1} \frac{3}{2})$

$t a n (s i n^{- 1} \frac{3}{5} + c o t^{- 1} \frac{3}{2})$

(M) Let $s i n^{- 1} \frac{3}{5} = x$ and cot^-1 $\frac{3}{2,}$ = y .

Then $s i n x = \frac{3}{5} c o t y = \frac{3}{2}$

Hence, tan x = $\frac{s i n x}{c o s x}$ = $\frac{\frac{3}{5}}{\frac{4}{5}}$ = $\frac{3}{4}$

$t a n (s i n^{- 1} \frac{3}{5} + c o t^{- 1} \frac{3}{2}) = t a n (x + y)$

$= \frac{t a n x + t a n y}{1 - t a n x t a n y}$ .

$\frac{\frac{3 \times 3 + 2 \times 4}{4 \times 3}}{\frac{4 \times 3 - 3 \times 2}{4 \times 3}}$

$\frac{4 \times 3 - 3 \times 2}{4 \times 3}$

$\frac{9 + 8}{12 - 6} = \frac{17}{6}$

37. $t a n^{- 1} (t a n \frac{7 π}{6})$

$t a n^{- 1} (t a n \frac{7 π}{6}) = t a n^{1} (t a n \frac{6 π + π}{6})$

$= t a n^{- 1} (t a n \frac{6 π}{6} + \frac{π}{6})$

$= t a n^{- 1} (t a n π + \frac{π}{6})$

$= t a n^{- 1} (t a n \frac{π}{6})$ { Ø tan ( $π$ + Ø ) = tan Ø as tan b (+) we in 3rd quadrant)}

$= \frac{π}{6}$ ∈ $(\frac{- π}{2}, \frac{π}{2})$

40. $c o s^{- 1} \frac{4}{5} + c o s^{- 1} \frac{12}{13} = c o s^{- 1} \frac{33}{65}$

$c o s^{- 1} \frac{4}{5} + c o s^{- 1} \frac{12}{13} = c o s^{- 1} \frac{33}{6 5^{°}} .$

Let cos^-1 $\frac{4}{5}$ and cos^-1 $\frac{12}{13}$ = y.

Then, $c o s x = \frac{4}{5} and cos y = \frac{12}{13} .$

Using $c o s (x + y) = c o s x c o s y - s i n x s i n y .$

$\Rightarrow c o s [c o s^{- 1} \frac{4}{5} + c o s^{- 1} \frac{12}{13}] = \frac{4}{5} \frac{12}{13} - \frac{3}{5} \times \frac{5}{13}$

$\Rightarrow c o s [c o s^{- 1} \frac{4}{5} + c o t^{- 1} \frac{12}{13}] = \frac{48 - 15}{65} = \frac{33}{65}$

$\Rightarrow c o s^{- 1} \frac{4}{5} + c o s^{- 1} \frac{12}{13} = c o s^{- 1} \frac{33}{65} . ?$

41. $c o s^{- 1} \frac{12}{13} + s i n^{- 1} \frac{3}{5} = s i n^{- 1} \frac{56}{65}$

$c o s^{- 1} \frac{12}{13} + s i n^{- 1} \frac{3}{5} = s i n^{- 1} \frac{56}{65}$

Let $c o s^{- 1} \frac{12}{13} = xandsin^{- 1} \frac{3}{5} = y .$

Thin, $c o s x = \frac{12}{13} and sin y = \frac{3}{5}$

Using $s i n (x + y) = s i n x c o s y + c o s x s i n y .$

$s i n [c o s^{- 1} \frac{12}{13} + s i n^{- 1} \frac{3}{5}] = \frac{5}{13} \times \frac{4}{5} + \frac{12}{13} \times \frac{3}{5} = \frac{20 + 36}{65} = \frac{56}{65} .$

$c o s^{- 1} \frac{12}{13} + s i n^{- 1} \frac{3}{5} = s i n^{- 1} \frac{56}{65} .$

43. $t a n^{- 1} \frac{1}{5} + t a n^{- 1} \frac{1}{7} + t a n^{- 1} \frac{1}{3} + t a n^{- 1} \frac{1}{8} = \frac{π}{4}$

LH.S $= (t a n^{- 1} \frac{1}{5} + t a n^{- 1} \frac{1}{7}) + (t a n^{- 1} \frac{1}{3} + t a n^{- 1} \frac{1}{8})$

$= t a n^{- 1} [\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \cdot \frac{1}{7}}] + t a n^{- 1} [\frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3}, \frac{1}{8}}] {\begin{matrix} ? u s i n g t a n^{- 1} x + t a n^{- 1} y & \frac{x + y}{1 - x y}, x y < 1 \end{matrix}}$

= $t a n^{- 1} [\frac{\frac{7 + 5}{7 \times 5}}{\frac{7 \times 5 - 1}{7 \times 5}}] + t a n^{- 1} [\frac{\frac{8 + 3}{5 \times 3}}{\frac{8 \times 3 - 1}{8 \times 3}}]$

$= t a n^{- 1} (\frac{12}{35 - 1}) + t a n^{- 1} (\frac{11}{24 - 1}) = t a n^{- 1} \frac{12}{34} + t a n^{- 1} \frac{11}{23}$

$= t a n^{- 1} \frac{6}{17} + t a n^{- 1} \frac{11}{23}$

$= t a n^{- 1} (\frac{\frac{6}{17} + \frac{1}{23}}{1 - \frac{6}{17} \times \frac{3}{23}}) = t a n^{- 1} (\frac{\frac{6 \times 23 + 11 \times 11}{17 \times 23}}{\frac{? 7 \times 23 - 6 \times 11}{17 \times 23}})$

$= t a n^{- 1} \frac{138 + 187}{391 - 66} = t a n^{- 1} \frac{325}{325} = t a n^{- 1} 1$

$= t a n^{- 1} (t a n \frac{1}{4})$

$= \frac{x}{4} =R.H.S$

51. $\begin{array}{l} s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}, t h e n x i s e q u a l t o \\ (A) 0, \frac{1}{2} (B) 1, \frac{1}{2} (C) 0 (D) \frac{1}{2} \end{array}$

$s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2} \to (1)$

(M) Let $x = s i n θ . T h e n, θ = s i n^{- 1} x .$

Putting this in $q^{n} (1)$ we get

$s i n^{- 1} (1 - x) - 2 \cdot θ = \frac{π}{2}$

$\Rightarrow s i n^{- 1} (1 - x) = \frac{π}{2} + 2 θ$

$\Rightarrow 1 - x = s i n (\frac{π}{2} + 2 θ) {s i n (\frac{π}{2} + x) = c o s x}$

$\Rightarrow 1 - x = c o s 2 θ$

$\Rightarrow 1 - x = 1 - 2 s i n^{2} θ \cdot {c o s 2 x = 1 - 2 s i n^{2} x}$

$\Rightarrow 1 - x = 1 - 2 x^{2} \cdot {s i n θ = x}$

$\begin{array}{l} \Rightarrow 2 x^{2} - x = 0 \\ \Rightarrow x (2 x - 1) = 0 \end{array}$

$\Rightarrow s o, x = 0 x 2 x - 1 = 0 x 2 x = 1 \to x = \frac{1}{2} .$

Putting $x = 0$ in qⁿ (1) .

L.H.S $= s i n^{- 1} (1 - 0) - 2 s i n^{- 1} 0 = s i n^{- 1} s i n \frac{x}{2} - 0 = \frac{π}{2} = R . H . S .$

$x = \frac{1}{2} q(1)$

$L.H.S= s i n^{- 1} (1 - \frac{1}{2}) - 2 s i n^{- 1} \frac{1}{2} = s i n^{- 1} (\frac{1}{2}) - 2 s i n^{- 1} \frac{1}{2}$

$= s i n^{- 1} \frac{1}{2} - 2 s i n^{- 1} \frac{1}{2}$

$= - s i n^{- 1} \frac{1}{2} = - s i n^{- 1} (s i n \frac{π}{6}) = - \frac{π}{6}$ $\neq$

So, $= 0 .$

Option (c) is correct.

21. Kindly Consider the following

Kindly go through the solution

35. Kindly Consider the following

Kindly go through the solution

47. Kindly Consider the following

Kindly go through the solution

NCERT Class 12 Math Chapter 2: Key Topics, Weightage

In the Inverse Trigonometry Class 12, students should practice the graphs, properties, and domain/range of inverse trigonometric functions. See below the topics covered in this chapter:

Exercise	Topics Covered
2.1	Introduction
2.2	Basic Concepts
2.3	Properties of Inverse Trigonometric Functions

Inverse Trigonometric Functions Class 12 Weightage in JEE Mains

Exam	Weightage
JEE Main	2% to 3%

Important Formulas of Inverse Trigonometry Class 12

Inverse Trigonometric Function Important Formulae for CBSE and Competitive Exams

Basic Properties

$\sin^{-1}(-x) = -\sin^{-1}(x), \quad x \in [-1, 1]$
$\cos^{-1}(-x) = \pi - \cos^{-1}(x), \quad x \in [-1, 1]$
$\tan^{-1}(-x) = -\tan^{-1}(x), \quad x \in \mathbb{R}$
$\cot^{-1}(-x) = \pi - \cot^{-1}(x), \quad x \in \mathbb{R}$
$\sec^{-1}(-x) = \pi - \sec^{-1}(x), \quad |x| \geq 1$
$\csc^{-1}(-x) = -\csc^{-1}(x), \quad |x| \geq 1$

Domain and Range of ITF

Function	Domain	Range
$\sin^{-1}x$	$[- 1, 1]$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$
$\cos^{-1}x$	$[- 1, 1]$	$[0, π]$
$\tan^{-1}x$	$(- \infty, \infty)$	$\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$
$\cot^{-1}x$	$(- \infty, \infty)$	$(0, π)$
$\sec^{-1}x$	$(- \infty, - 1] \cup [1, \infty)$	$[0, π] ∖ {\frac{π}{2}}$
$\csc^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \setminus \{0\}$

Sum and Difference Formulas

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad x \in [-1, 1]$
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$
$\sec^{-1}x + \csc^{-1}x = \pi, \quad |x| \geq 1$

NCERT Solutions for Inverse Trigonometric Functions- FAQs

Students can access few of the frequently asked questions below;

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Inverse Trigonometric Function Exercise-wise NCERT Solutions

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Inverse Trigonometric Function Exercise 2.1 Solutions

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NCERT Class 12 Math Chapter 2: Key Topics, Weightage

Inverse Trigonometric Functions Class 12 Weightage in JEE Mains

Important Formulas of Inverse Trigonometry Class 12

Inverse Trigonometric Function Important Formulae for CBSE and Competitive Exams

Basic Properties

Domain and Range of ITF

NCERT Solutions for Inverse Trigonometric Functions- FAQs

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