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prove that: (sec8A - 1)/(sec4A - 1) = tan8A/tan2A

prove that:


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

Grade:11

7 Answers

vikas askiitian expert
509 Points
13 years ago

(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

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

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

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

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.

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Sagar Singh

B.Tech, IIT Delhi

vikas askiitian expert
509 Points
13 years ago

(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

vikas askiitian expert
509 Points
13 years ago

(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

Krish Gupta
askIITians Faculty 82 Points
3 years ago
 

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.

 Hope it helps, thanks.

 

Krish Gupta
askIITians Faculty 82 Points
3 years ago
Given, (sec8A - 1) / (sec4A - 1) 
= (1/cos8A) - 1) / (1/cos4A) - 1
= (1 - cos8A)/cos8A) / (1 - cos4A) / cos4A) 
= cos4A (1 - cos8a) / (cos8A (1 - cos4A)) 
= cos4A(1 - (1 - 2sin²4A)) / cos8A (1 - (1 - 2sin²2A))
= cos4A sin²4A / (cos8A sin²2A) 
= (2 cos4A sin4A) sin4A / (2 cos8A sin²2A) 
= sin8A sin4A / (2 cos8A sin²2A) 
= tan 8A * (sin 4A / 2 sin^2 2A)
 = tan 8A * (cos 2A / sin 2A)
  = tan 8A/tan 2A
 
Hope this helps!
 

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