# Example 5 - Chapter 2 Class 12 Inverse Trigonometric Functions

Last updated at April 16, 2024 by Teachoo

Examples

Example 1
Important

Example 2

Example 3 (i) Important

Example 3 (ii)

Example 4 Important

Example 5 Important You are here

Example 6 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams

Question 6 Important Deleted for CBSE Board 2025 Exams

Question 7 Important Deleted for CBSE Board 2025 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

Last updated at April 16, 2024 by Teachoo

Example 5 Write cotβ1 (1/β(π₯^2 β 1)), |π₯| > 1 in the simplest form. cot-1 (1/β(π₯^2 β 1)) Putting x = sec ΞΈ = cotβ1 (1/β(γπ¬ππγ^πβ‘π β 1)) = cotβ1 (1/β(γ(π + γπππ§γ^πγβ‘π½ ) β 1)) = cotβ1 (1/β(tan^2β‘ΞΈ )) We write 1/β(π₯^2 β 1) in form of cot Whenever there is β(π₯^2β1) , we put x = sec ΞΈ = cotβ1 (1/tanβ‘ΞΈ ) = cotβ1 (cot ΞΈ) = ΞΈ We assumed x = sec ΞΈ sec ΞΈ = x ΞΈ = secβ1 x Hence, our equation becomes cotβ1 (1/β(π₯^2β1)) = ΞΈ cotβ1 (1/β(π₯^2β1)) = secβ1 x