Example 6 - Chapter 2 Class 12 Inverse Trigonometric Functions

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