**NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions**

NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions will give you an idea of the technics used in solving the different types of questions based on the inverse trigonometric functions. The NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions are solved and scrutinized by a team of experts. The purpose of these NCERT solutions is to help students in clearing their doubts so that they could achieve excellent marks in the exam.

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | Chapter 2-Trigonometric functions |

Chapter 8-Application of Integrals |

**NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions**

**Q1.3Sin ^{-1}x = Sin^{-1}(3x – 4x³) ; x ∈ [1/2,-1/2]**

Ans. Considering Sin^{-1}x = θ

sin θ = x

Taking RHS ,we have

Sin^{-1}(3x – 4x³)

Putting the value of x = sin θ

Sin^{-1}(3sin θ – 4sin ³θ)[since sin 3θ =3sin θ – 4sin ³θ]

Sin^{-1} (sin 3θ)

= 3θ = 3Sin^{-1}x = LHS

Hence proved

Q2.**3cos ^{-1}x = cos^{-1}(3x – 4x³) ; x ∈ [1/2,1]**

Ans. Considering cos^{-1}x = θ

cos θ = x

Taking RHS ,we have

cos^{-1}(3x – 4x³)

Putting the value of x = cos θ

cos^{-1}(3cos³ θ – 4cos θ)[since cos 3θ = 4cos³θ -3cos θ]

cos^{-1} (cos 3θ)

= 3θ = 3cos^{-1}x = LHS

Hence proved

**Q3. Prove that **

**tan ^{-1}2/11 + tan^{-1}7/24= tan^{-1}1/2**

Taking the LHS

tan^{-1}2/11 + tan^{-1}7/24

Applying the formula

= Tan^{-1}(125/250)

= Tan^{-1}1/2 = RHS, Hence proved

**NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions**

**Q4. Prove that**

**2tan ^{-1}(1/2) +tan^{-1}(1/7) =tan^{-1}(31/17)**

Taking the LHS

2tan^{-1}(1/2) +tan^{-1}(1/7)

Reawritting it as following

tan^{-1}(1/2)+tan^{-1}(1/2) +tan^{-1}(1/7)

Applying the formula

= tan^{-1}(1/2) +tan^{-1}(9/13)

Again applying the same formula

= tan^{-1}(31/17) = RHS, Hence proved

**Q5.Write the function in the simplest form**

Ans. Considering the value of x = tan θ

We get

= θ/2

**Q6. Write the function in the simplest form**

Ans. Considering x = cosec θ

Tan^{-1}(tan θ) = θ

Since x = cosec θ⇒θ = cosec^{-1}x

**Q7. Write the function in the simplest form**

Ans. The given function is

Since 1 – cos x = 2sin²(x/2) and 1 + cos x = 2cos²(x/2)

=tan^{-1}(tan x/2) = x/2

**NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions**

**Q8. Write the function in the simplest form**

Ans. The given function is

For writting the given function in the form of tan,dividing the numerator and denominator by cos x

Writing it in the following form

= tan^{-1 }[tan(π/4 – x)]

= π/4 – x

**Q9. Write the function in the simplest form**

Ans. The given function is

Considering x = a sin θ, we have

= tan^{-1 }(tan θ) = θ

Since, x = a sin θ, hence sin θ = x/a

θ = sin^{-1}(x/a)

**Q10. Write the function in the simplest form**

Ans.The given function is

Considering x = a tan θ, we have

= tan^{-1 }(tan 3θ) = 3θ

Since x = a tan θ ⇒ tan θ = x/a , θ = tan^{-1 }(x/a)

Therefore 3θ = 3tan^{-1 }(x/a)

Find the value of each of the following:

Ans.

= tan^{-1}1 ⇒ π/4

cot π/2 =0

**NCERT solutions for class 12 maths exercise 2.2 of chapter 2-Inverse Trigonometric functions**

**Q12. cot ( sin ^{-1}α + cos^{-1} α)**

Ans. We are given

cot ( sin^{-1}α + cos^{-1} α)

Since, sin^{-1}x + cos^{-1} x) = π/2

cot π/2 = 0

Ans. We are given the function

Considering x = tan A and y = tan B

= tan (A + B)

Since x = tan A and y = tan B

**Q14.If sin[ sin ^{-1}(1/5) + cos^{-1}x ] =1, then find the value of x**

Ans. We are given that

sin[ sin^{-1}(1/5) + cos^{-1}x ] =1

Substituting 1=Sin π/2

sin[ sin^{-1}(1/5) + cos^{-1}x ] =sin π/2

sin^{-1}(1/5) + cos^{-1}x = π/2

Since sin^{-1}A + cos^{-1}A= π/2

Therefore

cos^{-1}x = π/2 – sin^{-1}(1/5) =cos^{-1}1/5

x = 1/5

Ans. We are given that

Applying the following formula

2x² -4 = -3

2x² = 1⇒ x = ±1/√2

**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 science **

**CBSE Class 9-Question paper of science 2020 with solutions**

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

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