Question

prove that :- sec8A-1/sec4A-1=tan8A/Tan2A

Answer 1

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
 
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Answer 2

Answer:

HEY MATE FIND THE ANSWER BELOW

Step-by-step explanation:

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

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