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NCERT Solutions
Class 12
Maths
Chapter 2 Inverse Trigonometric Functions

Frequently Asked Questions

The chapter focuses on inverse trigonometric functions, their domains, ranges, and important identities.

Principal values help define a unique output for each input, making inverse functions well-defined.

The solutions show correct step-by-step methods and proper use of identities, helping to reduce common mistakes during exams.

You can find NCERT Solutions for Class 12 Maths Chapter 2 on the ALLEN website. The solutions cover the NCERT exercises on inverse trigonometric functions with step-by-step explanations.

NCERT Class 12 Maths Chapter 2 explains principal values by restricting the domain of trigonometric functions so that their inverse functions have unique values.

NCERT Class 12 Maths Chapter 2 includes Exercise 2.1 and Exercise 2.2. These exercises cover principal values, domain and range, evaluation, properties, and related expressions.

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NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Chapter 2 of Class 12 Maths, Inverse Trigonometric Functions builds on the basics of trigonometry and helps students understand how inverse functions work within a defined domain and range. This chapter focuses on inverse trigonometric functions, their principal values, domains, ranges, and standard identities. These concepts are important for solving higher-level problems in calculus and for applying trigonometry correctly in exams.

ALLEN offers Class 12 Math NCERT Solutions that are created in strict accordance with the most recent NCERT textbook and the most recent CBSE syllabus. Each question in Class 12 Maths Chapter 2-NCERT Solutions is explained step-by-step to assist students grasp topics without difficulty. Frequent practice of these problems increases students' confidence, accuracy, and readiness for competitive and board exams. Revision is made simple and efficient by the use of student-friendly language.

1.0Key Concepts of Class 12 Maths Chapter 2 Inverse Trigonometric Functions

Class 12 Maths Chapter 2, Inverse Trigonometric Functions, explains the concept of inverse trigonometric functions, their principal values, domains, ranges, and properties. The main concepts covered in NCERT Solutions Class 12 Maths Chapter 2 include:

  • Inverse Trigonometric Functions: Understand the meaning, notation, and basic properties of inverse trigonometric functions.
  • Principal Value Branches: Learn how suitable restrictions on the domain help define inverse trigonometric functions with unique values.
  • Domain and Range: Study the domain and range of functions such as sin⁻¹x, cos⁻¹x, and tan⁻¹x.
  • Graphs of Inverse Functions: Understand the basic graphical representation of inverse trigonometric functions.
  • Standard Identities: Learn important identities and properties used to solve questions based on inverse trigonometric functions.
  • Simplification of Expressions: Practise simplifying expressions by applying the properties and identities of inverse trigonometric functions.

2.0NCERT Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Detailed Solutions

EXERCISE - 2.1

Find the principal values of the following Q. 1 to 10.

  1. sin−1(2−1​) Sol. Let sin−1(2−1​)=θ⇒sinθ=−21​ We know that the range of principal value of sin−1x is [−2π​,2π​]. ⇒sinθ=−21​=−sin6π​=sin(−6π​)(∵sin(−θ)=−sinθ) ⇒θ=−6π​, where θ∈[−2π​,2π​] ⇒sin−1(−21​)=−6π​
  2. cos−1(23​​) Sol. Let cos−1(23​​)=θ⇒cosθ=23​​ We know that the range of principal value of cos−1x is [0,π]. ∴cosθ=23​​=cos6π​ ⇒θ=6π​, where θ∈[0,π] ⇒cos−1(23​​)=6π​
  3. cosec−1(2) Sol. Let cosec−1(2)=θ⇒cosecθ=2. We know that the range of principal value of cosec−1x is [−2π​,2π​]−{0}. ⇒cosecθ=2=cosec6π​ ⇒θ=6π​, where θ∈[−2π​,2π​]−{0} ⇒cosec−1(2)=6π​
  4. tan−1(−3​) Sol. Let tan−1(−3​)=θ⇒tanθ=−3​ We know that the range of principal value of tan−1x is (−2π​,2π​) ⇒tanθ=−3​=−tan3π​=tan(−3π​) (∵tan(−θ)=−tanθ) ⇒θ=−3π​, where θ∈(−2π​,2π​) ⇒tan−1(−3​)=−3π​
  5. cos−1(−21​) Sol. Let cos−1(−21​)=θ⇒cosθ=−21​ We know that the range of principal value of cos−1x is [0,π] ⇒cosθ=−21​=−cos3π​=cos(π−3π​)=cos32π​ (∵cos(π−θ)=−cosθ) ⇒θ=32π​; where θ∈[0,π] ⇒cos−1(−21​)=32π​
  6. tan−1(−1) Sol. Let tan−1(−1)=θ⇒tanθ=−1 We know that the range of principal value of tan−1x is (−2π​,2π​) ∴tanθ=−1=−tan4π​=tan(−4π​) (∵tan(−θ)=−tanθ) ⇒θ=−4π​; where θ∈(−2π​,2π​) ∴tan−1(−1)=−4π​
  7. sec−1(3​2​) Sol. Let sec−1(3​2​)=θ⇒secθ=3​2​ We know that the range of principal value of sec−1x is [0,π]−{2π​} ∴secθ=3​2​=sec(6π​) ⇒θ=6π​ where θ∈[0,π]−{2π​} ⇒sec−1(3​2​)=6π​
  8. cot−1(3​) Sol. Let cot−1(3​)=θ⇒cotθ=3​ We know that the range of principal value of cot−1x is (0,π) ∴cotθ=3​=cot6π​ ⇒θ=6π​; where θ∈(0,π) ⇒cot−1(3​)=6π​
  9. cos−1(−2​1​) Sol. Let cos−1(−2​1​)=θ⇒cosθ=−2​1​ We know that the range of principal value of cos−1x is [0,π] ∴cosθ=−2​1​=−cos4π​=cos(π−4π​) [∵cos(π−θ)=−cosθ] ⇒θ=43π​, where θ∈[0,π] ⇒cos−1(−2​1​)=43π​
  10. cosec−1(−2​) Sol. Let cosec−1(−2​)=θ⇒cosecθ=−2​ We know that the range of principal value of cosec−1x is [−2π​,2π​]−{0} ∴cosecθ=−2​=−cosec4π​=cosec(−4π​) (∵cosec(−θ)=−cosecθ) ⇒θ=−4π​, where θ∈[−2π​,2π​]−{0} ⇒cosec−1(−2​)=−4π​

Find the values of the following Q. 11 & 12:

  1. tan−1(1)+cos−1(−21​)+sin−1(−21​) Sol. Let tan−1(1)=x ⇒tanx=1=tan4π​⇒x=4π​ where principal value of x∈(−2π​,2π​) ∴tan−1(1)=4π​ Let cos−1(−21​)=y ⇒ cosy =−21​=−cos(3π​)=cos(π−3π​) =cos(32π​)(∵cos(π−θ)=−cosθ) ⇒y=32π​, where principal value of y∈[0,π] ∴cos−1(−21​)=32π​ Let sin−1(−21​)=z sinz=−21​=−sin(6π​)=sin(−6π​) (∵sin(−θ)=−sinθ)

⇒z=−6π​, where principal value of z∈[−2π​,2π​] ∴sin−1(−21​)=−6π​ ∴tan−1(1)+cos−1(−21​)+sin−1(−21​) =x+y+z=4π​+32π​−6π​ =123π+8π−2π​=129π​=43π​

  1. cos−1(21​)+2sin−1(21​)

Sol. Let cos−1(21​)=x ⇒cosx=21​=cos3π​ ⇒x=3π​∈[0,π] Again, let sin−1(21​)=y⇒siny=21​=sin6π​ ⇒y=6π​∈[−2π​,2π​] ∴cos−1(21​)+2sin−1(21​)=x+2y=3π​+(2×6π​)=32π​


Choose the correct answer in the following questions from 13 & 14.

  1. If sin−1x=y, then (A) 0≤y≤π (B) −2π​≤y≤2π​ (C) 0<y<π (D) −2π​<y<2π​

Sol. (B) As range of sin−1x is [−2π​,2π​], therefore −2π​≤y≤2π​.

  1. tan−13​−sec−1(−2) is equal to (A) π (B) −3π​ (C) 3π​ (D) 32π​

Sol. (B) Let tan−1(3​)=x ⇒tanx=3​⇒tanx=tan3π​ ⇒x=3π​∈(−2π​,2π​) Let sec−1(−2)=y⇒secy=−2 ⇒secy=−sec3π​ ⇒secy=sec(π−3π​)[∵sec(π−θ)=−secθ] ⇒secy=sec(32π​) ⇒y=32π​∈[0,π]−{2π​} ∴tan−1(3​)−sec−1(−2) =x−y=3π​−32π​=−3π​ Hence, the correct option is (B).


EXERCISE - 2.2

(Q. Nos. 1 to 2) Prove the following questions.

  1. 3sin−1x=sin−1(3x−4x3),x∈[−21​,21​]

Sol. Let sin−1x=θ⇒x=sinθ, then

 RHS ====​sin−1(3x−4x3)sin−1[3sinθ−4sin3θ]sin−1[sin3θ]=3θ[∵sin3θ=3sinθ−4sin3θ]3sin−1x=LHS​

  1. 3cos−1x=cos−1(4x3−3x),x∈[21​,1]

Sol. Let cos−1x=θ⇒x=cosθ, then

 RHS ​=cos−1(4x3−3x)=cos−1[4cos3θ−3cosθ]=cos−1[cos3θ]=3θ[∵cos3θ=4cos3θ−3cosθ]=3cos−1x= LHS ​


(Q Nos. 3 to 7) Write the following functions in the simplest form.

  1. tan−1(x1+x2​−1​),x=0 Sol. Put x=tanθ⇒θ=tan−1x

∴ and sinθ=2sin2θ​cos2θ​]∴​tan−1(x1+x2​−1​)=tan−1(tanθ1+tan2θ​−1​)=tan−1(tanθsec2θ​−1​)[∵1+tan2θ=sec2θ]=tan−1[tanθsecθ−1​]=tan−1[cosθsinθ​cosθ1​−1​]=tan−1[cosθsinθ​cosθ1−cosθ​​]=tan−1[cosθ1−cosθ​×sinθcosθ​]=tan−1[sinθ1−cosθ​]=tan−1[2sin2θ​cos2θ​2sin22θ​​][∵1−cosθ=2sin22θ​=tan−1[2cos2θ​2sin2θ​​]=tan−1[tan2θ​]=2θ​=2tan−1x​[ From equation (1)] tan−1(x1+x2​−1​)=21​tan−1x​

  1. tan−1(1+cosx1−cosx​​),0<x<π Sol. tan−1(1+cosx1−cosx​​)

​=tan−1​2cos2(x/2)2sin2(x/2)​​​( and 1+cosx=2cos2(2x​)∵1−cosx=2sin2(2x​)​)=tan−1(tan22x​​)=tan−1(tan2x​)=2x​​

  1. tan−1(cosx+sinxcosx−sinx​),4−π​<x<43π​ Sol. tan−1(cosx+sinxcosx−sinx​)=tan−1(cosxcosx​+cosxsinx​cosxcosx​−cosxsinx​​) (inside the bracket divide numerator and denominator by cosx )

=tan−1(1+tanx1−tanx​)=tan−1[tan(4π​−x)]=4π​−x

  1. tan−1(a2−x2​x​),∣x∣<a

Sol. Put x=asinθ

⇒∴​ax​=sinθ⇒sin−1(ax​)=θtan−1(a2−x2​x​)=tan−1(a2−a2sin2θ​asinθ​)=tan−1(a1−sin2θ​asinθ​)=tan−1(cosθsinθ​)(⇒cosx=1−sin2x​∵sin2x+cos2x=1​)=tan−1(tanθ)=θ=sin−1(ax​) [From equation (1)] ​

  1. tan−1(a3−3ax23a2x−x3​),a>0;3​−a​<x<3​a​

Sol. Put x=atanθ

⇒∴​ax​=tanθ⇒θ=tan−1(ax​)tan−1(a3−3ax23a2x−x3​)=tan−1[a3−3a(atanθ)23a2(atanθ)−(atanθ)3​]=tan−1[a3(1−3tan2θ)a3(3tanθ−tan3θ)​]=tan−1[1−3tan2θ3tanθ−tan3θ​]=tan−1(tan3θ)(∵tan3θ=1−3tan2θ3tanθ−tan3θ​)=3θ=3tan−1(ax​)[ From equation (1)]​


(Q.Nos 8 & 9) Find the values of each of the following questions.

  1. tan−1[2cos(2sin−121​)]

Sol.

​tan−1[2cos(2sin−121​)]=tan−1[2cos{2sin−1(sin6π​)}](∵sin6π​=21​)=tan−1[2cos(2×6π​)]=tan−1[2cos3π​]=tan−1[2×21​]=tan−1(1)(∵cos3π​=21​)=tan−1(tan4π​)=4π​(∵tan4π​=1)​

  1. tan21​[sin−1(1+x22x​)+cos−1(1+y21−y2​)], ∣x∣<1,y>0 and xy<1

Sol. tan21​[sin−1(1+x22x​)+cos−1(1+y21−y2​)] Put x=tanθ,y=tanϕ

⇒⇒⇒⇒​tan21​[sin−1(1+tan2θ2tanθ​)+cos−1(1+tan2ϕ1−tan2ϕ​)]tan21​[sin−1(sin2θ)+cos−1(cos2ϕ)][∵sin−1(sinx)=x,x∈[−2π​,2π​]]tan21​[2θ+2ϕ]tan[θ+ϕ]1−tanθ⋅tanϕtanθ+tanϕ​=1−xyx+y​​


(Q. Nos 10 to 12) Find the values of each of the expressions

  1. sin−1(sin32π​)

Sol. sin−1(sin32π​)=sin−1[sin(π−3π​)]

=sin−1(sin3π​)​=3π​∈[−2π​,2π​](∵sin−1(sinθ)=θ,θ∈[−2π​,2π​])​

  1. tan−1(tan43π​)

Sol. tan−1(tan43π​)=tan−1[tan(π−4π​)]

==​tan−1(−tan4π​)[∵−tanθ=tan(−θ)]tan−1[tan(−4π​)]=−4π​∈(−2π​,2π​)[∵tan−1(tanθ)=θ,θ∈(−2π​,2π​)]​

  1. tan(sin−153​+cot−123​) Sol. tan(sin−153​+cot−123​) Put sin−153​=θ⇒sinθ=53​,

class-12-maths-chapter-2-ques-12-sol

Now, tanθ=43​ and cot−123​=ϕ ⇒cotϕ=23​, Now tanϕ=32​ Since, tan(θ+ϕ)=1−tanθtanϕtanθ+tanϕ​

=[1−(43​×32​)43​+32​​]=617​

Choose the correct answer in the following questions from 13 to 15.

  1. cos−1(cos67π​) is equal to? (A) 67π​ (B) 65π​ (C) 3π​ (D) 6π​ Sol. (B) cos−1(cos67π​)=cos−1[cos(2π−65π​)]

∴cos−1(cos67π​)=​cos−1[cos(65π​)]=65π​∈[0,π][∵cos−1(cosθ)=θ,θ∈[0,π]]​

  1. sin[3π​−sin−1(−21​)] is equal to (A) 21​ (B) 31​ (C) 41​ (D) 1 Sol. (D) sin[3π​−sin−1(−21​)]

​=sin[3π​−sin−1(−sin6π​)][∵sin6π​=21​][∵sin(−x)=−sinx]=sin[3π​−sin−1{sin(−6π​)}]=sin[3π​−(−6π​)]=sin[3π​+6π​]=sin2π​=1​

  1. tan−1(3​)−cot−1(−3​) is equal to (A) π (B) −2π​ (C) zero (D) 23​ Sol. (B) tan−1(3​)−cot−1(−3​)

​=tan−1(3​)−[π−cot−1(3​)][∵cot−1(−x)=π−cot−1x]=tan−1(3​)−π+cot−1(3​)=3π​−π+6π​=−2π​​

MISCELLANEOUS EXERCISE

(Q. Nos 1 and 2) : Find the value of following functions:

  1. cos−1(cos613π​)

Sol.

cos−1(cos613π​)[∵cos−1(cosθ)=​=cos−1[cos(2π+6π​)]=cos−1[cos(6π​)]=6π​∈[0,π]θ,θ∈[0,π] and ∵cos(2π+θ)=cosθ]​

  1. tan−1(tan67π​)

Sol.

​tan−1(tan67π​)=tan−1[tan(π+6π​)][∵tan−1(tanθ)=θ,θ∈(−2π​,2π​) and ∵tan(π+θ)=tanθ]n−1(tan67π​)=tan−1[tan6π​]=6π​∈(−2π​,2π​)​

Prove the result of following functions:

  1. 2sin−1(53​)=tan−1(724​)

Sol.

​tan[2sin−1(53​)]=tan[tantan[2sin−1(53​)]=(724​)​

mis-exer-class-12-maths-chap-2-ques-2

sinθ=(53​)⇒tanθ=43​

Now,

LHS​=tan2θ=1−tan2θ2tanθ​=1−169​2×43​​=23​×716​=724​=RHS​

Hence Proved

  1. sin−1(178​)+sin−1(53​)=tan−1(3677​)

Sol.

​tan[sin−1(178​)+sin−1(53​)]=tan[tan−1(3677​)]tan[sin−1(178​)+sin−1(53​)]=(3677​)​

Let sin−1178​=θ⇒sinθ=178​⇒tanθ=158​ Again, let sin−153​=ϕ

⇒sinϕ=53​⇒tanϕ=43​

Now,

LHS​=tan(θ+ϕ)=1−tanθtanϕtanθ+tanϕ​​



ques-4-sol-class-12-maths-chap-2





mis-exer-ques-4-class-12-chap-2-maths



  1. cos−1(54​)+cos−1(1312​)=cos−1(6533​)

Sol.

​ Given, cos−1(54​)+cos−1(1312​)=cos−1(6533​)cos[cos−1(54​)+cos−1(1312​)]=cos[cos−1(6533​)]cos[cos−1(54​)+cos−1(1312​)]=6533​​

Let cos−154​=θ ⇒

cosθ=54​⇒sinθ=53​

ques-5-sol-chap-2-maths-class-12


[cosϕ=1312​]


ques-5-sol-maths-class-12-chap-2


[sinϕ=135​]

Again, Let cos−11312​=ϕ

 Now, LHS ⇒cos(θ​=cos(θ+ϕ)=cosθcosϕ−sinθsinϕ+ϕ)=(54​×1312​)−(53​×135​)=6548​−6515​=6533​= RHS ​

Hence proved.


  1. cos−1(1312​)+sin−1(53​)=sin−1(6556​) Sol. Given, cos−1(1312​)+sin−1(53​)=sin−1(6556​)

​sin[cos−1(1312​)+sin−1(53​)]=sin[sin−1(6556​)]sin[cos−1(1312​)+sin−1(53​)]=6556​​

Let cos−1(1312​)=θ

[cosθ=1312​]

mis-exer-ques-6-class-12-maths-chap-2


[cosθ=135​] Again, Let sin−153​=ϕ

[cosθ=53​]

ques-6-sol-class-12-maths-chap-2-mis-exer


[cosθ=54​]

 Now, LHS ​=sin(θ+ϕ)=sinθcosϕ+cosθsinϕ=(135​×54​)+(1312​×53​)=6520​+6536​=6556​= RHS  Hence proved. ​

  1. tan−1(1663​)=sin−1(135​)+cos−1(53​) Sol. We have sin−1135​+cos−153​=tan−11663​

⇒tan[sin−1135​+cos−153​]=1663​

Let [sin−1(135​)=θ]


ques-7-sol-class-12-maths-chap-2


= [sinθ=135​]

⇒tanθ=125​

Again, let cos−153​=ϕ

[cosθ=53​]

mis-exer-ques-7-sol-class-12-chap-2-maths


[tanθ=34​]

∵tan(θ+ϕ)=1−tanθ⋅tanϕtanθ+tanϕ​

tan(θ+ϕ)​=1−125​⋅34​125​+34​​=3636−20​3615+48​​=3616​3663​​=1663​= RHS  Hence proved. ​

  1. Prove that :

tan−1x​=21​cos−1(1+x1−x​),x∈[0,1]

Sol. Let tan−1x​=θ then x=tan2θ

 RHS ​=21​cos−1(1+x1−x​)=21​cos−1(1+tan2θ1−tan2θ​)[∵cos2θ=1+tan2θ1−tan2θ​]=21​cos−1(cos2θ)=θ=tan−1x​= LHS Hence Proved ​

  1. cot−1(1+sinx​−1−sinx​1+sinx​+1−sinx​​)=2x​,x∈(0,4π​) Sol. LHS =cot−1(1+sinx​−1−sinx​1+sinx​+1−sinx​​) Now, we can write

1+sinx​​=sin22x​+cos22x​+2sin2x​cos2x​​=(sin2x​+cos2x​)2​=​sin2x​+cos2x​​=sin2x​+cos2x​[∵2x​∈(0,8π​)]​

Similarly, we can get

1−sinx​​=​cos2x​−sin2x​​=cos2x​−sin2x​​

∵2x​∈(0,8π​) ⇒cos2x​−sin2x​>0 On substituting above two values in Equation (1), we get L.H.S. =

​cot−1((sin2x​+cos2x​)−(cos2x​−sin2x​))(sin2x​+cos2x​)+(cos2x​−sin2x​)​)=cot−1(2sin2x​2cos2x​​)=cot−1(cot2x​)=2x​= R.H.S. ​

  1. tan−1(1+x​+1−x​1+x​−1−x​​)=4π​−21​cos−1x,

−2​1​≤x≤1

L. H.S.

​=tan−1(1+x​+1−x​1+x​−1−x​​),−2​1​≤x≤1 Put x=cos2θ⇒21​cos−1x=θ=tan−1(1+cos2θ​+1−cos2θ​1+cos2θ​−1−cos2θ​​)=tan−1(2​∣cosθ∣+2​∣sinθ∣2​∣cosθ∣−2​∣sinθ∣​)​∵x2​=∣x∣∵−2​1​≤x≤1⇒−2​1​≤cos2θ≤1⇒0≤2θ≤43π​⇒0≤θ≤83π​∴cosθ>0 and sinθ>0​​=tan−1(cosθ+sinθcosθ−sinθ​)=tan−1(1+tanθ1−tanθ​)=tan−1(tan(4π​−θ))=4π​−θ=4π​−21​cos−1x= RHS ​


(Q. Nos 11 & 12) : Solve the following equations for x:

  1. 2tan−1(cosx)=tan−1(2cosecx) Sol. Given, 2tan−1(cosx)=tan−1(2cosecx)

tan[2tan−1(cosx)]=tan[tan−1(2cosecx)]

[on taking tan both sides]

Put tan−1(cosx)=θ⇒tanθ=cosx ⇒tan2θ=2cosecx ⇒1−tan2θ2tanθ​=sinx2​[∵tan2θ=1−tan2θ2tanθ​] ⇒1−cos2xcosx​=sinx1​[∵tanθ=cosx] ⇒sin2xcosx​=sinx1​ ⇒cotx=1 ⇒cotx=cot4π​⇒x=4π​

  1. tan−1(1+x1−x​)=21​tan−1x,(x>0)

Sol. Given; tan−1(1+x1−x​)=21​tan−1x ⇒2tan−1(1+x1−x​)=tan−1x ⇒tan{2tan−1(1+x1−x​)}=x ⇒tan2θ=x ⇒1−tan2θ2tanθ​=x [ Let tan−1(1+x1−x​)=θ⇒tanθ=1+x1−x​] ⇒1−[tan(tan−1(1+x1−x​))]22tan[tan−1(1+x1−x​)]​=x ⇒[(1+x)2(1+x)2−(1−x)2​(1+x)2(1−x)​​]=x ⇒[(1+x)2−(1−x)22(1−x)(1+x)​]=x ⇒[1+x2+2x−1−x2+2x2(1−x2)​]=x ⇒[4x2(1−x2)​]=x ⇒(2x(1−x2)​)=x ⇒2x1−x2​=x ⇒1−x2=2x2 ⇒1=3x2 ⇒x2=31​ ⇒x=±3​1​ ⇒x=3​1​ [∵x>0 given, so we do not take x=−3​1​] Choose the correct answer in the following questions from 13 to 15.

  1. sin(tan−1x);∣x∣<1 is equal to (A) 1−x2​x​ (B) 1−x2​1​ (C) 1+x2​1​ (D) 1+x2​x​

Sol. (D) Let tan−1x=θ or x=tanθ ⇒sinθ=cosecθ1​

​=1+cot2θ​1​=1+x21​​1​=1+x2​x​​

  1. If sin−1(1−x)−2sin−1x=2π​, then x is equal to (A) 0,21​ (B) 1,21​ (C) 0 (D) 21​ Sol. (C) We have, sin−1(1−x)−2sin−1x=2π​ ⇒sin−1(1−x)=2π​+2sin−1x ⇒sin{sin−1(1−x)}=sin(2π​+2sin−1x) [Let sin−1x=θ⇒x=sinθ ]

⇒1−x=cos(2sin−1x) ⇒1−x=cos2θ ⇒1−x=1−2sin2θ ⇒1−x=(1−2x2) ⇒x=2x2 ⇒x(2x−1)=0 ⇒x=0,21​ For, x=21​, we obtain

LHS​=sin−1(1−x)−2sin−1x=sin−121​−2sin−121​=−sin−121​=−6π​=RHS​

So, x=21​ is not a root of the given equation. Clearly, x=0 satisfies the given equation. Hence, x=0 is a root of the given equation, the correct option is (C)

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for Class 12 Maths arranged chapter-wise, with exercise-wise answers, useful formulas, and detailed steps to make concepts easier to understand and questions easier to solve.

Chapter Number

Chapterwise NCERT Solutions

Chapter 1

Relations and Functions

Chapter 3

Matrices

Chapter 4

Determinants

Chapter 5

Continuity and Differentiability

Chapter 6

Application of Derivatives

Chapter 7

Integrals

Chapter 8

Application of Integrals

Chapter 9

Differential Equations

Chapter 10

Vector Algebra

Chapter 11

Three Dimensional Geometry

Chapter 12

Linear Programming

Chapter 13

Probability

4.0Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise-wise Solutions

Exercise

Number of Questions

Important Topics

Exercise 2.1 Solutions

14 Questions & Solutions

Domain, range, principal values and evaluation of inverse trigonometric functions

Exercise 2.2 Solutions

15 Questions & Solutions

Properties of inverse trigonometric functions and evaluation of related expressions

5.0Key Features and Benefits of Class 12 Maths Chapter 2 Inverse Trigonometric Functions

  • Solutions are prepared according to the latest NCERT syllabus for Class 12 Maths followed by CBSE schools.
  • Step-by-step explanations make it easier to understand the method used to solve each question.
  • Solutions cover the exercises and questions given in the NCERT Class 12 Maths textbook.
  • Regular practice with NCERT questions can help students improve their calculation speed, accuracy, and time management.
  • A strong understanding of inverse trigonometric functions, their properties, and identities can support preparation for mathematics olympiads and other competitive examinations.

Table of Contents

  • 1.0Key Concepts of Class 12 Maths Chapter 2 Inverse Trigonometric Functions
  • 2.0NCERT Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Detailed Solutions
  • 2.1EXERCISE - 2.1
  • 2.2EXERCISE - 2.2
  • 2.3MISCELLANEOUS EXERCISE
  • 3.0Class 12 Maths NCERT Solutions – Chapter-wise Links
  • 4.0Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise-wise Solutions
  • 5.0Key Features and Benefits of Class 12 Maths Chapter 2 Inverse Trigonometric Functions