Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
Close
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

⇐ BackProceed to Checkout ⇒
There are no items in this cart.
Continue Shopping

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

prove that:


(sec8A - 1)/(sec4A - 1) = tan8A/tan2A

Grade:11

7 Answers

vikas askiitian expert
509 Points
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

SAGAR SINGH - IIT DELHI
879 Points
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.

 

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

SAGAR SINGH - IIT DELHI
879 Points
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

 

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

vikas askiitian expert
509 Points
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

vikas askiitian expert
509 Points
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

Krish Gupta
askIITians Faculty 82 Points
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.

 

Krish Gupta
askIITians Faculty 82 Points
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!
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Trigonometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Register ICAT

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More