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Example 5 Write cotβˆ’1 (1/√(π‘₯^2 βˆ’ 1)), |π‘₯| > 1 in the simplest form. cot-1 (1/√(π‘₯^2 βˆ’ 1)) Putting x = sec ΞΈ = cotβˆ’1 (1/√(γ€–π¬πžπœγ€—^πŸβ‘π›‰ βˆ’ 1)) = cotβˆ’1 (1/√(γ€–(𝟏 + γ€–π­πšπ§γ€—^πŸγ€—β‘πœ½ ) βˆ’ 1)) = cotβˆ’1 (1/√(tan^2⁑θ )) We write 1/√(π‘₯^2 βˆ’ 1) in form of cot Whenever there is √(π‘₯^2βˆ’1) , we put x = sec ΞΈ = cotβˆ’1 (1/tan⁑θ ) = cotβˆ’1 (cot ΞΈ) = ΞΈ We assumed x = sec ΞΈ sec ΞΈ = x ΞΈ = secβˆ’1 x Hence, our equation becomes cotβˆ’1 (1/√(π‘₯^2βˆ’1)) = ΞΈ cotβˆ’1 (1/√(π‘₯^2βˆ’1)) = secβˆ’1 x

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo