prove that:
(sec8A - 1)/(sec4A - 1) = tan8A/tan2A
prove that:
(sec8A - 1)/(sec4A - 1) = tan8A/tan2A
paradox xyz cool , 14 Years ago
Grade 11
Trigonometry> plz solve dis problem...
7 Answers
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)
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
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.
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All the best.
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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.
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.
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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=1-2SIN24ACOS4A =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
Last Activity: 5 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
Last Activity: 5 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!
Hope this helps!
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