Question

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

Answer 1

Step-by-step explanation:

(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