(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
										

										
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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Sagar Singh
B.Tech, IIT Delhi






										

										
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






										

										
(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
										

										

(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