Ex 2.2, 6 - Chapter 2 Class 12 Inverse Trigonometric Functions

Transcript

Ex 2.2, 6 Write the function in the simplest form: tan−1 1/√(𝑥^2−1), |x| > 1 tan−1 (1/√(𝑥^2 − 1)) Putting x = sec θ = tan−1 (1/√(sec^2⁡θ − 1)) = tan−1 (1/√(〖(1 + tan^2〗⁡θ ) − 1)) = tan−1 (1/√(〖1 −1 + tan^2〗⁡θ )) = tan−1 (1/√(tan^2⁡θ )) We write 1/√(𝑥^2 − 1) in form of tan Whenever there is √(𝑥^2−1) , we put x = sec θ (sec2 θ = 1 + tan2 θ) = tan−1 (1/tan⁡θ ) = tan−1 (cot θ) = tan−1 tan (90 – θ) = 90 – θ = 𝜋/2 – θ We assumed x = sec θ sec θ = x θ = sec-1 x (1/tan⁡𝜃 " = cot θ" ) (cot θ = tan (90 – θ) ) Hence, tan-1 (1/√(𝑥^2−1)) = 𝜋/2 – θ = 𝝅/𝟐 – sec−1 x