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paradox xyz cool Grade: 11
        

prove that:


(sec8A - 1)/(sec4A - 1) = tan8A/tan2A

6 years ago

Answers : (5)

vikas askiitian expert
510 Points
										

(SEC8A -1)/SEC4A-1 = TAN8A/TAN2A


SECX =1/COSX


SO


    LHS = (1-COS8A)COS4A/(1-COS4A)COS8A


    COS8A=1-2SIN24A    


    COS4A =1-2SIN22A


AFTER PUTTING THESE


  LHS = (SIN24A)COS4A/(SIN22A)COS8A


        = (2SIN4ACOS4A)SIN4A/2SIN22ACOS8A                             (multiplying dividing by 2)


2SIN4ACOS4A =COS8A


 SO


    LHS =(SIN8A)SIN4A/(2SIN22A)COS8A =(TAN8A)SIN4A/2SIN22A


 SIN4A = 2SIN2ACOS2A


SO


  LHS =(TAN8A)COS2A/SIN2A=TAN8A/TAN2A = RHS


 HENCE PROVED

6 years ago
SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


L.H.S. = (sec 8A -1) / (sec 4A -1)
=> [(1 - cos 8A)/ cos 8A] / [(1 - cos 4A)/ cos 4A]
=> [2 sin² 4A / 2 sin² 2A] * [cos 4A / cos 8A]
=> [(2 sin 4A * cos 4A) * sin 4A / cos 8A] / [2 sin² 2A]

=> [(sin 8A / cos 8A) * sin 4A] / [2 sin² 2A]
=> tan 8A * (sin 4A / 2 sin² 2A)
=> tan 8A * (2sin 2A * cos 2A / 2 sin² 2A)
=> tan 8A * ( cos 2A / sin 2A)
=> tan 8A * cot 2A

=> tan 8A / tan 2A ==> R.H.S.


 








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Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi







6 years ago
SAGAR SINGH - IIT DELHI
879 Points
										

Dear student,


L.H.S. = (sec 8A -1) / (sec 4A -1)
=> [(1 - cos 8A)/ cos 8A] / [(1 - cos 4A)/ cos 4A]
=> [2 sin² 4A / 2 sin² 2A] * [cos 4A / cos 8A]
=> [(2 sin 4A * cos 4A) * sin 4A / cos 8A] / [2 sin² 2A]

=> [(sin 8A / cos 8A) * sin 4A] / [2 sin² 2A]
=> tan 8A * (sin 4A / 2 sin² 2A)
=> tan 8A * (2sin 2A * cos 2A / 2 sin² 2A)
=> tan 8A * ( cos 2A / sin 2A)
=> tan 8A * cot 2A

=> tan 8A / tan 2A ==> R.H.S


 








Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.


All the best.


Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.


Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..


 


Askiitians Expert


Sagar Singh


B.Tech, IIT Delhi







6 years ago
vikas askiitian expert
510 Points
										

(SEC8A -1)/SEC4A-1 = TAN8A/TAN2A


 


SECX =1/COSX SO


 


LHS = (1-COS8A)COS4A/(1-COS4A)COS8A


COS8A=1-2SIN24A


COS4A =1-2SIN22A


AFTER PUTTING THESE


 


LHS = (SIN24A)COS4A/(SIN22A)COS8A


 


      = (2SIN4ACOS4A)SIN4A/2SIN22ACOS8A                                                  (multiplying dividing by 2)


 


     we have 2SIN4ACOS4A =SIN8A


 


SO LHS =(SIN8A)SIN4A/(2SIN22A)COS8A


 


          =(TAN8A)SIN4A/2SIN22A


 


we have SIN4A= 2SIN2ACOS2A


 


SO LHS =(TAN8A)COS2A/SIN2A=TAN8A/TAN2A = RHS


HENCE PROVED

6 years ago
vikas askiitian expert
510 Points
										


(SEC8A -1)/SEC4A-1 = TAN8A/TAN2A


SECX =1/COSX


SO


  LHS = (1-COS8A)COS4A/(1-COS4A)COS8A


  COS8A=1-2SIN24A   


COS4A =1-2SIN22A


AFTER PUTTING THESE


LHS = (SIN24A)COS4A/(SIN22A)COS8A   


      = (2SIN4ACOS4A)SIN4A/2SIN22ACOS8A                             (multiplying dividing by 2)


so


   2SIN4ACOS4A =SIN8A          




LHS =(SIN8A)SIN4A/(2SIN22A)COS8A =(TAN8A)SIN4A/2SIN22A


SIN4A = 2SIN2ACOS2A


so


  LHS =(TAN8A)COS2A/SIN2A=TAN8A/TAN2A = RHS


 hence proved

6 years ago
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