prove that

i. (1 + sin2A - cos2A)/(1 + sin2A + cos2A) = tanA

ii. (sinA + sin2A)/(1 + cosA + cos2A) = tanA

iii. 2cosA = [2(2 + 2cos4A)^1/2]^1/2

iv. 1 + cos56 + cos 58 - cos66 = 4cos28.cos29.sin33

1) (1 + SIN2A - COS2A)/(1+SIN2A + COS2A) = TANA

SIN2A =2TANA/1+TAN²A      &  COS2A =1-TAN²A/1+TAN²A

PUTTING THESE

 LHS =(2TAN²A + 2TANA)/2TANA+2

       =TANA = RHS

2)   (SINA + SIN2A)/(1+COSA+COS2A)    =   TANA

 COS2A = 2COS²A - 1       &   SIN2A =2SINACOSA

AFTER PUTTING THESE

LHS = (SINA+2SINACOSA)/COSA+2COS²A

      =SINA(1+2COSA) / COSA(1+2COSA)

      =SINA/COSA=TANA =RHS

HENCE PROVED