Example 5 - Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
Not clear how to approach
Not clear how to approach
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