# Maths XII chapter 2 inverse trigonometric functions ex 2 1

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Chapter 2 Mathematics XII Exercise 2.1 WWW.TIWARIACADEMY.COM (www.tiwariacademy.com) (Class XII) Exercise 2.1 Question 1: www.tiwariacademy.com 1 Find the principal value of −1 (− ) 2 Answer 1: 1 Let −1 (− ) = , Then 2 1 = − = − ( ) = (− ) 2 6 6 π π We know that the range of the principal value branch of sin −1 is [- , ] and 2 2 π 1 sin (- ) = 6 2 1 π Therefore, the principal value of sin-1 (- ) is - . 2 6 www.tiwariacademy.com Question 2: √3 Find the principal value of −1 ( ) 2 Answer 2: www.tiwariacademy.com √3 Let −1 ( ) = , then 2 √3 = ( ) 2 6 We know that the range of the principal value branch of cos −1 is [0, π] and cos = π cos ( ) = 6 √3 2 √3 π Therefore, the principal value of cos -1 ( ) is . 2 6 1 Free web support in Education (www.tiwariacademy.com) (Class XII) Question 3: www.tiwariacademy.com Find the principal value of cosec−1 (2). Answer 3: π Let cosec−1 (2) = y. Then, cosec y = 2 = cosec ( ) 6 π π We know that the range of the principal value branch of cosec −1 is [- 2 , 2 ] -{0} π and cosec ( ) = 2. 6 π Therefore, the principal value of cosec−1 (2) is . 6 Question 4: www.tiwariacademy.com Find the principal value of −1 (−√3). Answer 4: π π Let tan-1 (-√3) = y, then tan y = -√3 = -tan = tan (- ) 3 3 π π We know that the range of the principal value branch of tan−1 is (- , ) and 2 2 π tan (- ) = -√3 3 π Therefore, the principal value of tan-1 (-√3) is - . 3 Question 5: www.tiwariacademy.com 1 Find the principal value of −1 (− ). 2 Answer 5: 1 1 π π 2π Let cos -1 (- ) = y, then cos y = - = -cos = cos (π- ) = cos ( ) 2 2 3 3 3 We know that the range of the principal value branch of cos −1 is [0, π] and 2π 1 cos ( ) = 3 2 1 Therefore, the principal value of cos -1 (- ) is 2 2π 3 . 2 Free web support in Education (www.tiwariacademy.com) (Class XII) Question 6: www.tiwariacademy.com Find the principal value of tan−1 (−1). Answer 6: π π Let tan−1 (−1) = y. Then, tan y = -1 = -tan ( ) = tan (- ) 4 4 π π We know that the range of the principal value branch of tan −1 is (- , ) and 2 2 π tan (- ) = -1 4 π Therefore, the principal value of tan−1 (−1) is - . 4 Question 7: www.tiwariacademy.com 2 Find the principal value of −1 ( ). 3 √ Answer 7: 2 2 √ √ Let −1 ( ) = , then sec = 3 = ( ) 3 6 We know that the range of the principal value branch of sec −1 is [0, ] − { } 2 and ( ) = 6 2 √3 . 2 Therefore, the principal value of −1 ( ) is . 3 6 √ Question 8: www.tiwariacademy.com Find the principal value of −1 √3. Answer 8: π Let cot -1 √3 = y, then cot y = √3 = cot ( ). 6 We know that the range of the principal value branch of cot −1 is (0, π) and π cot ( ) = √3. 6 π Therefore, the principal value of cot -1 √3 is . 6 3 Free web support in Education (www.tiwariacademy.com) (Class XII) Question 9: www.tiwariacademy.com Find the principal value of −1 (− 1 ). √2 Answer 9: Let cos -1 (- cos y = - 1 ) = y, then √2 1 √ π π 3π = -cos ( ) = cos (π- ) = cos ( ). 2 4 4 4 We know that the range of the principal value branch of cos −1 is [0, π] and 3π cos ( ) = 4 1 . √2 Therefore, the principal value of cos -1 (- 1 ) is √2 3π 4 . Question 10: Find the principal value of −1 (−√2). Answer 10: Let −1 (−√2) = , then = −√2 = − ( ) = (− ) 4 4 π π We know that the range of the principal value branch of cosec−1 is [- , ] -{0} 2 2 π and cosec (- ) = -√2. 4 π Therefore, the principal value of cosec -1 (-√2) is - . 4 4 Free web support in Education (www.tiwariacademy.com) (Class XII) Question 11: www.tiwariacademy.com 1 1 Find the value of −1 (1) + −1 (− ) + −1 (− ). 2 2 Answer 11: Let −1 (1) = , then tan = 1 = 4 We know that the range of the principal value branch of tan−1 is (− , ) . 2 2 ∴ −1 (1) = 4 1 Let −1 (− ) = , then 2 1 2 cos = − = − = ( − ) = ( ) 2 3 3 3 −1 We know that the range of the principal value branch of cos is [0, ]. 1 ∴ −1 (− ) = 2 2 3 1 Let −1 (− ) = , then 2 1 sin = − = − = (− ) 2 6 6 π π We know that the range of the principal value branch of sin−1 is [- , ]. 2 2 1 π ∴ sin-1 (- ) = 2 6 Now, 1 1 tan-1 (1) + cos -1 (- ) + sin-1 (- ) 2 2 = 2 3 + 8 − 2 9 3 + − = = = 4 3 6 12 12 4 5 Free web support in Education (www.tiwariacademy.com) (Class XII) Question 12: www.tiwariacademy.com 1 1 Find the value of −1 ( ) + 2−1 ( ) 2 2 Answer 12: 1 1 π Let cos -1 ( ) = x, then cos x = = cos 2 2 3 We know that the range of the principal value branch of cos−1 is [0, π]. 1 ∴ cos -1 ( ) = 2 π 3 1 1 π Let sin-1 (- ) = y, then sin y = = sin 2 2 6 π π We know that the range of the principal value branch of sin−1 is [- , ]. 2 2 ∴ 1 −1 ( ) 2 = 6 Now, 1 1 2 −1 ( ) + 2−1 ( ) = + 2 × = + = . 2 2 3 6 3 3 3 Question 13: If sin−1 x = y, then 2 2 2 2 (A) 0 ≤ ≤ (B) − ≤ ≤ (C) 0 < < (D) − < < Answer 13: www.tiwariacademy.com It is given that sin−1 x = y. π π We know that the range of the principal value branch of sin−1 is [- , ]. 2 2 π π 2 2 Therefore, - ≤ y ≤ . Hence, the option (B) is correct. 6 Free web support in Education (www.tiwariacademy.com) (Class XII) Question 14: www.tiwariacademy.com −1 √3 − −1 (−2) is equal to (A) π (C) (B) − (D) 3 3 2 3 Answer 14: Let −1 √3 = , then tan = √3 = 3 π π We know that the range of the principal value branch of tan−1 is (- , ). 2 2 π ∴ tan-1 √3 = 3 Let sec -1 (-2) = y, then sec y = -2 = -sec π π 2π = sec (π- ) = sec ( ) 3 3 3 π We know that the range of the principal value branch of sec−1 is [0, π]- { } 2 ∴ sec -1 (-2) = 2π 3 Now, tan-1 √3-sec -1 (-2) = π 2π π =3 3 3 Hence, the option (B) is correct. www.tiwariacademy.com 7 Free web support in Education

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