Question

Prove that
sec8A-1/sec 4A - 1=cot 2A/cot8A​

Answer 1

Answer:

Step-by-step explanation:

L.H.S:

sec8A - 1 / sec4A - 1

=> [1 - cos8A / cos8A]  /  [1 - cos4A/cos4A]

=> [cos4A/cos8A] [ 1 - cos8A / 1 - cos4A ]

//remember cos2A = cos²A - sin²A = 1 - 2sin²A

=> [cos4A/cos8A] [ 1 - (1 - 2sin²4A) / 1 - (1 - 2sin²2A)]

=> [cos4A/cos8A] [ 2sin²4A / 2sin²2A]

=> (2sin4A.cos4A) (sin4A) / (2sin2A.cos8A)(sin2A)

//remember 2sinAcosA = sin2A

=> sin8Asin4A / (2sin²2A.cos8A)

=> (sin8A/cos8A) (sin4A/2sin²2A)

//remember sin2A = 2sinAcosA

=> tan8A * (2sin2Acos2A/2sin²2A)

=> tan8A * (cos2A/sin2A)

=> tan8A * cot2A

//remember cotA = 1/TanA

=> cot2A/cot8A

=> R.H.S

Hence proved.