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
Last Activity: 14 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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Askiitians Expert
Sagar Singh
B.Tech, IIT Delhi
vikas askiitian expert
Last Activity: 14 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
Last Activity: 14 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)