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

10 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

10 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.

All the best.

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Sagar Singh

B.Tech, IIT Delhi

10 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

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.

Sagar Singh

B.Tech, IIT Delhi

10 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

10 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

one year 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.

one year 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!