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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
(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
LHS =(SIN8A)SIN4A/(2SIN22A)COS8A =(TAN8A)SIN4A/2SIN22A
SIN4A = 2SIN2ACOS2A
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. 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. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. 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. 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. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
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.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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. 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. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar.. Askiitians Expert Sagar Singh B.Tech, IIT Delhi
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
(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
SECX =1/COSX SO
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
(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
(SEC8A -1)/SEC4A-1 = TAN8A/TAN2A SECX =1/COSX SO LHS = (1-COS8A)COS4A/(1-COS4A)COS8A COS8A=1-2SIN24A
so
2SIN4ACOS4A =SIN8A
hence proved
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.
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.
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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