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NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.1 Question 1 to 14 Answer: Downlaod pdf

Find the principal values of the following: 1. sin1 1 2 2. cos1 3 2 3. cosec1 (2) 4. tan1 ( 3) 5. cos1 1 2 6. tan1 (1) 7. sec1 2 3 8. cot1 ( 3) 9. cos1 1 2 10. cosec1 ( 2 ) Find the values of the following: 11. tan1(1) + cos1 1 2 + sin1 1 2 12. cos1 1 2 + 2 sin1 1 2 13. If sin1 x = y, then (A) 0 y p (B) 2 2 y p p (C) 0 < y < p (D) 2 2 y p p < < 14. tan1 ( ) 1 3 sec 2 is equal to (A) p (B) 3 p (C) 3 p (D) 2 3 p

ex.2.1 Class 12 Math Chapter 2 ncert solutions

Other EXERCISE for Class 12 Chapter 2 Inverse Trigonometric Functions NCERT Solutions




Find the principal values of the following:

Question 1:sin–1 1 2   −

ncert class 12 maths chapter 2 Inverse Trigonometric Functions

Question 2:cos–1 3 2  

exercise 2.1 maths class 12 Chapter 2 Inverse Trigonometric Functions

Question 3:cosec–1 (2)

cbse class 12 maths ncert solutions Chapter 2

Question 4:tan–1 (− 3)

Class 12 maths solutions Inverse Trigonometric Functions

Question 5:cos–1 1 2   −

ncert class 12 chemistry chapter 2 exercise solutions

Question 6:tan–1 (–1)

Inverse Trigonometric Functions Class 12 ncert solutions

Question 7:sec–1 2 3     

Class 12 Maths NCERT Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2.1

Question 8:cot–1 ( 3)

Exercise 2.1 class 12 Math ncert solutions Chapter 2

Question 9:cos–1 1 2   −   

Find the principal values of the following: 1. s

Question 10:cosec–1 ( − 2 ) Find the values of the following:

class 12 maths ncert solution pdf download Chapter 2

Question 11:tan–1(1) + cos–1 1 2 − + sin–1 1 2 −

Inverse Trigonometric Functions Math ncert solution class 12

Question 12:cos–1 1 2 + 2 sin–1 1 2

Inverse Trigonometric Functions ncert solutions class 12

Question 13:If sin–1 x = y, then (A) 0 ≤ y ≤ π (B) 2 2 y π p − ≤ ≤ (C) 0 < y < π (D) 2 2 y π p − < <

CBSE NCERT Solutions For Class 12 Inverse Trigonometric Functions

Question 14:tan–1 ( ) 1 3 sec 2 − − − is equal to (A) π (B) 3 p − (C) 3 p (D) 2 3 p

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