Question

(sec8A-1)÷(sec4A-1)=sec8A÷tan2A

Answer 1

Answer:

(sec8A - 1) ÷ (sec4A - 1)  

we know that secx = 1/ cosx

= (1/cos8A) - 1) ÷ (1/cos4A) - 1

= (1 - cos8A)/cos8A) ÷ (1 - cos4A) / cos4A)  

= cos4A (1 - cos8a) / (cos8A (1 - cos4A))  

we know that cos2A = 1- 2sin²A

= cos4A(1 - (1 - 2sin²4A)) ÷ cos8A (1 - (1 - 2sin²2A))

= cos4A(1 - 1 + 2sin²4A)) ÷ cos8A (1 - 1 + 2sin²2A))

= 2cos4A sin²4A ÷  2(cos8A sin²2A)

= cos4A sin²4A ÷ (cos8A sin²2A)  

= (2 cos4A sin4A) sin4A / (2 cos8A sin²2A)  

we know that sin2A = 2 sinA.CosA

= sin8A sin4A / (2 cos8A sin²2A)  

= tan 8A × (sin 4A / 2 sin^2 2A)

= tan 8A × (cos 2A / sin 2A)

= tan 8A ×cot2A

 = tan 8A ÷tan 2A

Step-by-step explanation: