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Ex 2.2, 5 - Simplify: tan-1 (root (1 + x2) - 1)/x - Chapter 2

Ex 2.2, 5 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Ex 2.2, 5 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3 Ex 2.2, 5 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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Transcript

Ex 2.2, 5 Write the function in the simplest form: tan−1 (√(1 + x^2 ) − 1)/x , x ≠ 0 tan−1 (√(1 + x^2 ) − 1)/x Putting x = tan 𝜃 = tan−1 ((√(𝟏 + 〖𝐭𝐚𝐧〗^𝟐 𝛉 )− 1)/(tan θ)) = tan−1 ((√(〖𝐬𝐞𝐜〗^𝟐 𝜽 ) − 1)/(tan θ)) = tan−1((sec⁡θ − 1)/(tan θ)) We write (√(1 + x^2 ) − 1)/x in form of tan Whenever there is √(1+ 𝑥^2 ) we put x = tan θ (sec2 θ = 1 + tan2 θ) = tan−1 ((1/cos⁡𝜃 − 1)/(sin⁡𝜃/cos⁡𝜃 )) = tan−1 (((1 − cos⁡θ)/cos⁡θ )/(sin⁡𝜃/cos⁡𝜃 )) = tan−1 ((1 −〖 cos〗⁡𝜃)/sin⁡𝜃 ) Using sin 2θ = 2 sin θ cos θ Replacing θ with 𝜃/2 sin 2θ/2 = 2 sin θ/2 cos θ/2 sin θ = 2 sin 𝜃/2 cos 𝜃/2 Also, cos 2θ = 1 – 2 sin2 θ Replacing θ with 𝜃/2 cos 2(𝜃/2) = 1 − 2 sin2 𝜃/2 cos θ = 1 − 2 sin2 𝜃/2 2 sin2 𝜃/2 = 1 – cos θ 1 – cos θ = 2 sin2 𝜃/2 So, our equation becomes = tan−1 ((𝟐 𝒔𝒊𝒏𝟐 𝜽/𝟐)/(𝟐 〖𝐬𝐢𝐧 〗⁡〖𝜽/𝟐〗 𝒄𝒐𝒔⁡〖 𝜽/𝟐〗 )) = tan−1 (〖sin 〗⁡〖𝜃/2〗/cos⁡〖 𝜃/2〗 ) = tan−1 (𝑡𝑎𝑛 𝜃/2) = 𝛉/𝟐 We assumed that x = tan θ θ = tan-1x Hence, our equation becomes tan−1 (√(1 + x^2 ) − 1)/x = θ/2 = 𝟏/𝟐 tan-1x

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.