Plus Two Math's Solution Ex 2.1 Chapter2 Inverse Trigonometric Functions

Here is the solution for Exercise 2.1 Chapter 2 Inverse Trigonometric Functions of NCERT plus two maths. Here we have given a detailed explanation of each and every exercise so that students can understand the concepts easily without any difficulty. The solution to each and every question is provided here so you can solve them by yourself if you don’t get the answer here.

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Board	SCERT, Kerala
Text Book	NCERT Based
Class	Plus Two
Subject	Math's Textbook Solution
Chapter	Chapter 2
Exercise	Ex 2.1
Chapter Name	Inverse Trigonometric Functions
Category	Plus Two Kerala

Kerala Syllabus Plus Two Math's Textbook Solution Chapter  2 Inverse Trigonometric Functions Exercises 2.1

Chapter  2  Inverse Trigonometric Functions Solution

• Exercises 2.1
• Exercises 2.2
• Miscellaneous Exercise Chapter 2

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  2  Inverse Trigonometric Functions Exercise   2.1

Question 1:

Find the principal values of the following:

(i)  (ii)  (iii) 

(iv)  (v)  (vi)  

(vii)  (viii)  (ix) 

(x) 

Solution 1:

(i) Let sin^-1 (-1/2) = y . then sin y=-1/2 = -sin(Π/6) = sin (-Π/6)

Implies the range of the principal value branch of sin^-1 is [-Π/2,Π/2] 

and sin (-Π/6) = -1/2 implies the principal value of sin^-1(-1/2)= -Π/6

(ii) 

Let cos^-1(√3/2) = y . then cosy=√3/2 = cos(Π/6) 

Implies the range of the principal value branch of  cos^-1 is [0,Π] 

and cos(Π/6) = √3/2 implies the principal value of cos^-1(√3/2)= Π/6

(iii)

Let cosec^-1(2) = y . then cosec y=2 = cosec(Π/6) 

Implies the range of the principal value branch of  cosec^-1 is  [-Π/2,Π/2]-{0} 

and cosec(Π/6) = 2 implies the principal value of cosec^-1(2)= Π/6

(iv)

Let tan^-1(-√3) = y . then tan y=-√3 = -tan(Π/3) =tan(-Π/3)

Implies the range of the principal value branch of  tan^-1 is  (-Π/2,Π/2)

and tan(-Π/3) = -√3 implies the principal value of tan^-1(-√3)= -Π/3

(v)

Let cos^-1(-1/2) = y . then cosy=-1/2 = -cos(Π/3) =cos(Π-Π/3) =cos(2Π/3)

Implies the range of the principal value branch of  cos^-1 is [0,Π] 

and cos(2Π/3) = -1/2 implies the principal value of cos^-1(-1/2)= Π/6

(vi)

Let tan^-1(-1) = y . then tan y=-1 = -tan(Π/4) =tan(-Π/4)

Implies the range of the principal value branch of  tan^-1 is  (-Π/2,Π/2)

and tan(-Π/4) = -1 implies the principal value of tan^-1(-1)= -Π/4

(vii)

Let sec^-1(2/√3) = y . then sec y=2/√3 = sec(Π/6) 

Implies the range of the principal value branch of  cosec^-1 is  [-0,Π]-{Π/2} 

and sec(Π/6)  = 2/√3 implies the principal value of sec^-1(2/√3)= Π/6

(viii)

Let cot^-1(√3) = y . then cot y=√3 = cot(Π/6) 

Implies the range of the principal value branch of  cot^-1 is  (0,Π)

and cot(Π/6) = √3 implies the principal value of cot^-1(√3)= Π/6

(ix)

Let cos^-1(-1/√2) = y . then cosy=-1/√2 = -cos(Π/4) =cos(Π-Π/4) =cos(3Π/4)

Implies the range of the principal value branch of  cos^-1 is [0,Π] 

and cos(3Π/4) = -1/√2 implies the principal value of cos^-1(-1/√2)= 3Π/4

(x) 

Let cosec^-1(-√2) = y . then cosec y=-√2 = -cosec(Π/4) =cosec(-Π/4) 

Implies the range of the principal value branch of  cosec^-1 is  [-Π/2,Π/2]-{0} 

and cosec(-Π/4) = -√2 implies the principal value of cosec^-1(-√2)= -Π/4

Question 2:

Find the values of the following:

1. 

2. 

Solution 2:





Question 3:

 is equal A)  B)  C)  D)  

 

 

 

Solution 3:

Let tan. Then tan x 

The range of the principle value branch of  is 



Let  Then, sec y= 

We know that the range of the principal value branch of  is 



Hence  = 

Hence option B

Question 4:

 If  then, A)  B) 

C) 0 < y < π D)  

Solution 4:

It is given that sin⁻¹x = y.

We know that the range of the principal value branch of sin⁻¹ is 

Therefore,.

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Chapter 2: Inverse Trigonometric Functions EX 2.1 Solution

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Plus Two Math's Chapter Wise Textbook Solution PDF Download

            Plus Two Math's Part I

• Chapter 1: Relations and Functions
• Chapter 2: Inverse Trigonometric Functions
• Chapter 3: Matrices
• Chapter 4: Determinants
• Chapter 5: Continuity and Differentiability
• Chapter 6: Application of Derivatives



             Plus Two Math's Part II

• Chapter  7: Application of Integrals
• Chapter  8: Integrals
• Chapter  9: Differential Equations
• Chapter 10: Vector Algebra
• Chapter 11: Three Dimensional Geometry
• Chapter 12: Linear Programming
• Chapter 13: Probability

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