prove that:

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

(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.

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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Askiitians Expert

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.

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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

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