What is the derivative of `sqrt((1+cosx)/(1-cosx))` ?
What is the derivative of `sqrt((1+cosx)/(1-cosx))` ?

Ex 9.3

Ex 9.3, 2

Ex 9.3, 3

Ex 9.3, 4 Important

Ex 9.3, 5

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Ex 9.3, 7 Important

Ex 9.3, 8

Ex 9.3, 9 Important

Ex 9.3, 10 Important

Ex 9.3, 11 Important

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Ex 9.3, 15 Important

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Ex 9.3, 17 Important

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Ex 9.3, 19 Important

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Ex 9.3, 22 Important

Ex 9.3, 23 (MCQ)

Last updated at Aug. 11, 2023 by Teachoo

Ex 9.3, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑𝑥=(1 − cos𝑥)/(1 + cos𝑥 ) 𝑑𝑦/𝑑𝑥 = (1 − cos𝑥)/(1 + cos𝑥 ) We know that cos 2x = 2cos2 x − 1 Putting x = 𝑥/2 cos 2𝑥/2 = 2 cos2 𝑥/2 − 1 cos x = 2 cos2 𝑥/2 − 1 1 + cos x = 2cos2 𝑥/2 We know cos 2x = 1 − 2sin2 x Putting x = 𝑥/2 cos2 2𝑥/2 = 1 − 2 sin2 𝑥/2 cos x = 1 − 2sin2 𝑥/2 1 − cos x = 2sin2 𝑥/2 Putting values in equation 𝒅𝒚/𝒅𝒙 = (𝟐 〖〖𝐬𝐢𝐧〗^𝟐 〗〖𝒙/𝟐〗 )/(𝟐 〖〖𝐜𝐨𝐬〗^𝟐 〗〖𝒙/𝟐〗 ) 𝑑𝑦 = (〖sin^2 〗〖𝑥/2〗 )/cos^(2 )〖𝑥/2〗 𝑑𝑥 𝑑𝑦 = tan2 𝑥/2 𝑑𝑥 Putting tan2 𝑥/2 = sec2 𝑥/2 − 1 𝒅𝒚 = (“sec2 ” 𝒙/𝟐 ” − 1″ )𝒅𝒙 Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x − 1 Putting x = 𝑥/2 tan2 𝒙/𝟐 = sec2 𝒙/𝟐 − 1 ∫1▒𝒅𝒚 = ∫1▒(𝒔𝒆𝒄𝟐 𝒙/𝟐−𝟏) 𝒅𝒙 y = ∫1▒〖𝑠𝑒𝑐2 𝑥/2 𝑑𝑥− ∫1▒𝑑𝑥〗 y = 𝟏/(𝟏/𝟐) tan 𝒙/𝟐 − x + C y = 2 tan 𝑥/2 − x + c y = 2 tan 𝑥/2 − x + C y = 𝟐 tan 𝒙/𝟐 − x + C is the general solution

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