**NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions**

NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions gives you an idea about a new trigonometric function which is known as an inverse trigonometric function, trigonometric functions like sin,cos, tan etc are defined as the ratio between the sides of a right triangle,if we are given the sides of right triangle ΔABC then we can easily get the value of tan A, sin B, etc.If we are given the ratio between the sides then how can we calculate the ∠A and ∠B if it is given that ∠C is 90°. For solving this problem the concept of opposite trigonometric function is raised known as Inverse Trigonometry Functions ,if sin A = 3/4 then A =sin^{-1}(3/4),we can see the value of A in the trigonometric table or evaluate with the help of the calculator.

All NCERT solutions of class 12 maths exercise 2.1 of chapter 2-Inverse Trigonometry Functions are solved by an expert of maths by a step by step method,so every student can understand the solutions easily.

**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 8-Application of Integrals |

Find the principal values of the following.

Ans.considering the given function equivalent to y

sin y = -1/2

sin π/6 = 1/2

sin y= -sin π/6

sin y = sin(-π/6)

y = -π/6

We know the range of Sin^{-1 }is [-π/2,π/2]

So, the principal value of y is ( -π/6)

Let y = cos^{-1}(√3/2)

cos y = √3/2

cos y = cos π/6

y = π/6

The range of principal value of cos^{-1 }is [0, π]

So, the required principal value is π/6

Let y = cosec^{-1}(2)

cosec y = 2

since cosec π/6 = 2

cosec y = cosec π/6

We know the range of Sin^{-1 }is [-π/2,π/2]

Therefore the principal value of cosec^{-1}(2) is π/6

Let y = tan^{-1}(√3)

tan y = √3

tan y = tan π/6

The range of principal value of tan^{-1 }is [-π/2,π/2]

Therefore the principal value of tan^{-1 }√3 is π/6

Let y = cos^{-1 }(-1/2)

cos y = -1/2

since cos π/3 = 1/2

cos y = – cos π/3 = cos (π-π/3)

cos y = cos (2π/3)

We, know the range of principal value of cos^{-1 }(-1/2) is [0,π]

Hence, the principal value of cos^{-1 }(-1/2) is 2π/3

Let y = tan^{-1 }(-1)

tan y = -1

since tan π/4 = 1

tan y = – tan π/4

tan y = tan (- π/4)

We, know the range of principal value of tan^{-1 } is [-π/2,π/2]

Hence, the principal value of tan^{-1 }(-1) is -π/4

Let y = sec^{-1 }(2/√3)

sec y = 2/√3

since sec π/6 = 2/√3

sec y = sec π/6

We, know the range of principal value of sec^{-1 } is [0,π]

Hence, the principal value of sec^{-1 }(-1/2) is π/6

Let y = cot^{-1 }(√3)

cot y = √3

since cot π/6 = √3

cot y = cot π/6

We, know the range of principal value of cot^{-1 } is [-π/2,π/2]

Hence, the principal value of cot^{-1 }(√3) is π/6

Let y = cos^{-1 }(-1/√2)

cos y = -1/√2

since cos π/4 = 1/√2

cos y = – cos π/4 = cos (π-π/4)

cos y = cos (3π/4)

We, know the range of principal value of cos^{-1 } is [0,π]

Hence, the principal value of cos^{-1 }(-1/2) is 3π/4

Let y = cosec^{-1}(-√2)

cosec y = -√2

since cosec π/4 = √2

cosec y = -cosec π/4 =cos(-π/4)

We know the range of cosec^{-1 }is [-π/2,π/2]

Therefore the principal value of cosec^{-1}(√2) is -π/4

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