Ex 2.2

Ex 2.2,1
Deleted for CBSE Board 2022 Exams

Ex 2.2, 2 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 3 Deleted for CBSE Board 2022 Exams

Ex 2.2, 4 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 5 Important Deleted for CBSE Board 2022 Exams You are here

Ex 2.2, 6 Deleted for CBSE Board 2022 Exams

Ex 2.2, 7 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 8 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 9 Deleted for CBSE Board 2022 Exams

Ex 2.2, 10 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 11 Deleted for CBSE Board 2022 Exams

Ex 2.2, 12 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 13 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 14 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 15 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 16 Deleted for CBSE Board 2022 Exams

Ex 2.2, 17 Deleted for CBSE Board 2022 Exams

Ex 2.2, 18 Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 20 (MCQ) Important Deleted for CBSE Board 2022 Exams

Ex 2.2, 21 (MCQ) Deleted for CBSE Board 2022 Exams

Chapter 2 Class 12 Inverse Trigonometric Functions (Term 1)

Serial order wise

Last updated at May 12, 2021 by Teachoo

Hello! We hope that the questions explained by Teachoo are helping you for your Board exams. If Teachoo has been of any help to you, would you consider making a donation to support us? Please click on this link to support us.

Hello! We hope that the questions explained by Teachoo are helping you for your Board exams. If Teachoo has been of any help to you, would you consider making a donation to support us? Please click on this link to support us.

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((secθ − 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/cos𝜃 − 1)/(sin𝜃/cos𝜃 )) = tan−1 (((1 − cosθ)/cosθ )/(sin𝜃/cos𝜃 )) = tan−1 ((1 −〖 cos〗𝜃)/sin𝜃 ) 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