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

```
9 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
```
9 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

```
9 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

```
9 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
```
9 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
```
9 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.
```
3 months 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!
```
3 months ago
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