NCERT Solutions for Class 12 Maths Exercise 2.2

Exercise 2.2 of class 12th Maths has 6 examples (Examples 3, 4, 5, 6, 7, 8) and 21 questions. All the examples and questions of this exercise are essential to do. But examples 3, 5, 6 and questions 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15, 18, 19, 20, 21 of exercise 2.2 of class 12th Maths are the most important examples and questions for the board exams because these examples and questions have already been asked in the exams several times.

If students can give 2 hours per day to exercise 2.2 of class 12th Maths, they need 3-4 days to complete exercise 2.2 of class 12th Maths. This time is an approximate time. This time can vary because no students can have the same working speed, efficiency, capability, etc.

Exercise 2.2 of class 12th Maths is not easy and not complex. It lies in the mid of easy and tough because some examples and questions of this exercise are easy, and some are complicated. However, the difficulty level of any topic/question varies from child to child. So, exercise 2.2 of class 12th Maths is easy, or tough depends on children also. Some children find it difficult, some find it easy, and some find it in the middle of easy and difficult.

In exercise 2.2 of class 12th Maths, students will learn properties of inverse trigonometric functions and solve questions with the help of these properties.
Properties of inverse trigonometric functions are:

• 1. (i) sin⁻¹(1/x) = cosec⁻¹x, x ≥ 1 or x ≤ – 1
  (ii) cos⁻¹(1/x) = sec⁻¹x, x ≥ 1 or x ≤ – 1
  (iii) tan⁻¹(1/x) = cot⁻¹x, x > 0
• 2. (i) sin⁻¹(–x) = – sin⁻¹x, x ∈ [– 1, 1]
  (ii) tan⁻¹(–x) = – tan⁻¹x, x ∈ R
  (iii) cosec⁻¹(–x) = – cosec⁻¹x, | x | ≥ 1
• 3. (i) cos⁻¹(–x) = π – cos⁻¹x , x ∈ [– 1, 1]
  (ii) sec⁻¹(–x) = π – sec⁻¹x, | x | ≥ 1
  (iii) cot⁻¹(–x) = π – cot⁻¹x, x ∈ R
• 4. (i) sin⁻¹x+ cos⁻¹x = π/2 , x ∈ [– 1, 1]
  (ii) tan⁻¹x+ cot⁻¹x = π/2, x ∈ R
  (iii) cosec⁻¹x+ sec⁻¹x = π/2, | x | ≥ 1
• 5. (i) tan⁻¹x + tan⁻¹y = tan⁻¹(x + y/1 – xy) , xy < 1 (ii) tan⁻¹x – tan⁻¹y = tan⁻¹(x – y/1 + xy) , xy > – 1
  (iii) tan⁻¹x+ tan⁻¹y = π + tan⁻¹(x + y/1 – xy), xy > 1; x, y > 0
• 6. (i) 2 tan⁻¹x = sin⁻¹(2x/1+ x²) , | x | ≤ 1
  (ii) 2 tan⁻¹x = cos⁻¹(1- x²/1+ x²), x ≥ 0
  (iii) 2 tan⁻¹x = tan⁻¹(2x/1-x²), – 1 < x < 1