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Misc 11 - Chapter 2 Class 12 Inverse Trigonometry - tan-1

Misc. 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Misc. 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3
Misc. 11 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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Misc 11 Prove tan−1 ((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) = π/4 − 1/2 cos-1 x, −1/√2 ≤ x ≤ 1 [Hint: Put x = cos 2θ] Solving L.H.S. tan−1 ((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) Putting x = cos 2𝜽 = tan−1 ((√(1 + cos 2θ) − √(1 − cos 2θ))/(√(1 + cos 2θ) + √(1 − cos 2θ)))Using cos 2θ = 2 cos2 θ − 1 1 + cos 2θ = 2 cos2 θ Also cos 2θ = 1 – 2 sin2 θ 2 sin2 θ = 1 − cos 2θ 1 − cos 2θ = 2sin2 θ = tan−1 ((√(𝟐 〖𝐜𝐨𝐬〗^𝟐 𝛉) − √(𝟐 〖𝐬𝐢𝐧〗^𝟐 𝛉))/(√(𝟐 〖𝐜𝐨𝐬〗^𝟐 𝛉) + √(𝟐 〖𝐬𝐢𝐧〗^𝟐 𝛉))) = tan−1 ((√2 cos θ − √2 sin θ)/(√2 cos θ + √(2 ) sin θ)) = tan−1 (( √2 (cos θ − sin θ))/(√2 (cos θ + sin θ))) = tan−1 (( (𝐜𝐨𝐬 𝛉 − 𝐬𝐢𝐧 𝛉))/( (𝐜𝐨𝐬 𝛉 + 𝐬𝐢𝐧 𝛉))) Dividing by cos θ = tan−1 ((cos⁡𝜃/cos⁡𝜃 − sin⁡𝜃/cos⁡𝜃 )/(cos⁡𝜃/cos⁡𝜃 + sin⁡𝜃/cos⁡𝜃 )) = tan−1 ((𝟏 − 𝐭𝐚𝐧 𝛉)/(𝟏 + 𝐭𝐚𝐧 𝛉)) We write ( (cos θ + sin θ))/( (cos θ − sin θ)) in form of tan We know that tan (x – y) = 𝑡𝑎𝑛⁡〖𝑥 −〖 𝑡𝑎𝑛〗⁡〖𝑦 〗 〗/(1 + 𝑡𝑎𝑛⁡〖𝑥 𝑡𝑎𝑛⁡𝑦 〗 ) So, we divide whole equation by cos = tan−1 ((𝟏 − tan θ)/(1 + 𝟏 . tan θ)) = tan−1 ((𝒕𝒂𝒏⁡〖 𝝅/𝟒〗 − tan⁡θ)/(1 + 𝐭𝐚𝐧⁡〖 𝝅/𝟒〗.〖 tan〗⁡θ )) = tan−1 (tan (𝜋/4−θ)) = 𝝅/𝟒 − 𝛉 We know that x = cos 2𝜃 Using tan (x – y ) = 𝒕𝒂𝒏⁡〖𝒙 −〖 𝒕𝒂𝒏〗⁡〖𝒚 〗 〗/(𝟏+ 𝒕𝒂𝒏⁡〖𝒙 𝒕𝒂𝒏⁡𝒚 〗 ) Replace x by 𝜋/4 and y by θ (As tan 𝜋/4 = 1 ) cos−1 x = 2θ 1/2 cos−1 x = θ Hence, tan−1 ((√(1 + x) − √(1 − x))/(√(1 + x) + √(1 − x))) = 𝜋/4 – θ = 𝝅/𝟒 – 𝟏/𝟐 cos−1x Hence proved

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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.