Please solve question 24 Please solve question 24 Cos (cos- x + cos- x + sin- = cos (cos¯ x + n/ 2) = — Sin (cos¯ x) -sin 24 Prove that : — cosec — tan TION We have, x) 24 Share with your friends Share 1 Shrutina Agarwal answered this Dear Student, 12tan-1αβ = xtan-1αβ = 2xαβ = tan 2xperpendicular = αbase = βhypotheneus = α2 + β2cos 2x = βα2 + β21 - 2sin2x = βα2 + β22sin2x = 1 - βα2 + β22sin2x = α2 + β2 - βα2 + β2sinx = ±12×α2 + β2 - βα2 + β2 = ±α2 + β2 - β2α2 + β2cosec x =±2α2 + β2α2 + β2 - βx = cosec-1±2α2 + β2α2 + β2 - βSo, cosec212tan-1αβ = coseccosec-1±2α2 + β2α2 + β2 - β2= 2α2 + β2α2 + β2 - β (1)12tan-1βα = ytan-1βα = 2yβα = tan 2yperpendicular = βbase = αhypotheneus = α2 + β2Now, cos 2y = αα2 + β22cos2y - 1 = αα2 + β22cos2y = αα2 + β2 + 12cos2y = α + α2 + β2α2 + β2 cosy = ± α + α2 + β22α2 + β2 secy =± 2α2 + β2α + α2 + β2 y = sec-1±2α2 + β2 α+ α2 + β2 So, sec212tan-1αβ = secsec-1±2α2 + β2α + α2 + β2 2= 2α2 + β2 α + α2 + β2 (2)LHS,α32cosec212tan-1αβ + β32sec212tan-1βα = α322α2 + β2α2 + β2 - β + β322α2 + β2 α + α2 + β2=α3α2 + β2α2 + β2 - β + β3α2 + β2 α2 + β2 + α = α3(α2 + β2) +α4α2 + β2 +β3(α2 + β2) - β4α2 + β2 α2 + β22 - αβ - βα2 + β2 + αα2 + β2= (α3+β3)(α2 + β2) +(α4 - β4)α2 + β2 α2 + β2 - αβ +α2 + β2 (α- β)= (α3+β3)(α2 + β2) +(α2 + β2) (α2 - β2)α2 + β2 α2 + β2 - αβ +α2 + β2 (α- β)= (α2 + β2)(α3+β3 + (α2 - β2)α2 + β2)α2 + β2 - αβ +α2 + β2 (α- β)= (α2 + β2)((α+β)(α2 + β2 - αβ) + (α - β)(α + β)α2 + β2)α2 + β2 - αβ +α2 + β2 (α- β)= (α2 + β2)(α+β)((α2 + β2 - αβ) + (α - β)α2 + β2)α2 + β2 - αβ +α2 + β2 (α- β)= (α2 + β2)(α+β) Hence proved Regards, 2 View Full Answer Suruchi Yadav answered this Please find this answer -1
